Here is an post-mortem analysis / "how to" for the Final. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may appear. Look for two pairs of similar systems to compare that have minimal differences in structure. If a compound is named, draw it out. If a reaction is involved, identify the type of reaction and then what the controlling factors are.

Qu1:
First identify the alpha positions, the positions adjacent to the carbonyl groups. C-H systems in the alpha positions are enolisable H. In the diagram below, the C-H in the alpha positions are shown in blue:

Qu2:
The reaction is electrophilic aromatic substitution, nitration and we need to look at the substituent effects on the aromatic ring. The substituents are a methoxy group, a ketone connected via the carbonyl C=O and a methyl group. The methoxy group is a strong electron donor which is strongly activating due to resonance effects. The ketone is attached through the carbonyl group so it is a moderate electron withdrawing group, a moderate deactivator due to resonance effects. The methyl (alkyl) group is a weak electron donor which is weakly activating due to inductive / hyperconjugation effects. So the reactivity:

Qu3:
The reactivity of carbonyl groups in carboxylic acid derivatives (acid chloride & ester) and aldehydes towards the Grignard reagent (a good nucleophile) is determined by the electronic effect of the changing substituents (blue bold) on each of the carbonyls. Electron withdrawing groups on the carbonyl make the carbon more electrophilic (and hence more reactive). -Cl is electronegative and electron withdrawing, -H is neutral (reference, does not donate or withdraw) and -OR is strongly electron donating. Hence, in terms of reactivity towards CH₃MgBr, acid chlorides are more reactive than aldehydes which in turn are more reactive than esters, i.e. we have:

Qu4:
Acidity...Remember the lower the pKa the stronger the acid, and think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A- (or know the pKas). Phenol is the most acidic (pKa = 10, H attached to electronegative O atom, conjugate base stabilised by the electronegative atom and resonance). Cyclopentadiene (pKa = 16) when deprotonated gives a conjugate base that is aromatic (a 6 pi system, hence stabilised). The terminal alkyne (pKa = 25) is more acidic than a simple alkane due to the hybridisation (sp) of the C atom with the H attached. So in terms of acidity...

Qu5:
The reaction of alkenes with HCl is electrophilic addition and it is controlled by the stability of the carbocation intermediate and follows Markovnikov's rule. The conjugated diene gives a resonance stabilised tertiary carbocation (most stable), 2-methylbut-2-ene gives a tertiary carbocation and but-2-ene gives a secondary carbocation (least stable) so the alkene reactivity order is:

based on stability

Qu6:
Basicity...Remember the lower the pKa the stronger the acid (hence weaker base), and think of the simple acid equation HA <=> H+    A- (where A- is the base we are thinking about) look for factors that affect the electron availability of A-....Here we are looking carbonyl systems. Two simple carbonyls and one dicarbonyl (active methylene compounds have more resonance stabilisation, the electrons are less available, they are less basic). In the ester enolate there is resonance stabilisation of the carbanion with the carbonyl group moving the -ve charge to the electronegative O atom (ester pKa about 25). In the dicarbonyl, we have a 1,3-dialdehyde. First let's compare a simple ester to a simple aldehyde. An ester is similar to an aldehyde, but there is competing resonance from the alkoxy O atom in the ester so there is less ability to stabilise the -ve charge, so esters are less acidic than aldehydes (ester pKa about 25, aldehyde about 17). Hence in the 1,3-dicarbonyl system, the dialdehyde is better stabilised by resonance and hence it is less basic than the simple ester. The final system is a carboxylate (i.e. carboxylic acid pKa = 5) where the -ve charge is stabilised due to the resonance with the carbonyl that allows the -ve charge to be delocalised across two electronegative O atoms. Therefore, in terms of basicity:

Qu7:
The reactivity of carbonyl groups of aldehydes and ketones towards hydrate formation is just an example of their characteristic nucleophilic additions. Higher reactivity means more hydrate formation and higher equilibrium constant,K, for that process. The electronic and steric factors of each of the substituents on each of the carbonyl groups need to be considered. In the simple aldehyde we have a neutral H atom (i.e. no electronic effect and small). In a simple ketone, here with two methyl groups, where the alkyl groups are weak electron donors. Electron donors on the carbonyl make the carbon less electrophilic. In contrast, in the 3rd system we have a -CF₃ group where the electronegative F atoms make it electron withdrawing which increases the carbonyl reactivity because it makes the carbonyl C more electrophilic. Hence in terms of reactivity towards water for the hydrate formation, aldehydes are more reactive than ketones. So, overall we have:

Qu8:
Basicity...Remember the lower the pKa the stronger the acid (hence weaker base), and think of the simple acid equation HA <=> H+    A- (where A- is the base we are thinking about) look for factors that affect the electron availability of A-.... Here we are looking at substituted anilines (i.e. substituted aromatic amine) systems. Electron donors (e.g. -OR) will donate electrons onto the aromatic ring helping make the N atom lone pair more available (i.e. more basic). Electron withdrawing groups (where nitro is stronger than a C=O system) will reduce electron availability and make the aniline less basic. Therefore, in terms of basicity:

AROMATICITY and RESONANCE:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria for the pi system (cyclic, planar, conjugated pi system with 4n+2 pi electrons).

Qu9:
Conjugation requires a continuous pi system that spans at least 3 atoms, so a simple alkene fits the criteria (pi system spans two atoms).

Qu10:
To be non-aromatic as drawn but with an aromatic resonance structure, we need to find a system that satisfies all the criteria for aromaticity except the electron count and we can therefore add / subtract electrons via resonance. Perhaps most critically, it means that all the atoms in the cyclic structure must already contribute to the p system. If "n" is not 1, then the Huckel rule, 4n + 2, will be 2 (n=0) or 10 (n=2) pi electrons... in this case a 2 pi electron system.

Qu11:
Given the set of compounds in the question, the most acidic is going to be from a cationic species (so it becomes neutral) and from 351 recall that -OH are more acidic than -NH.

Qu12:
To be non-aromatic as drawn, we need to find a system that fails on one of the first three criteria (i.e. it lacks a cyclic conjugated pi system) but has a conjugate base that satisfies all four of the criteria for aromaticity (i.e. it has a pi system that is cyclic, planar, fully conjugated and contains 4n+2 pi electrons). The most likely scenario will involve loss of H+ from a sp3 center to create a conjugated lone pair (so adding 2 electrons to the pi system) ... for clarity the aromatic resonance structures of the conjugate bases are also shown:

or

Qu13:
The most common form of tautomerism switches H and double bond positions... To be non-aromatic as drawn, we need to find a system that fails on one of the first three criteria (i.e. it lacks a cyclic conjugated pi system) but has a tautomer that satisfies all four of the criteria for aromaticity (i.e. it has a pi system that is cyclic, planar, fully conjugated and contains 4n+2 pi electrons):

Qu14:
In order to be anti-aromatic, we need a system that meets the first 3 criteria but has 4n pi electrons (an even number of pi electron pairs). In this case assuming the molecule is planar (as stated in the question), there then would be 2 C=C = 4 pi electrons (even number of pi electron pairs):

Qu15:
Heterocycles have a non-carbon atom in the cyclic system (i.e. all structures in the 3rd row of the answer set). Of these heterocycles, only two are uncharged (i.e. neutral). The initial base form needs to be aromatic, so we need to find a system that satisfies all the four criteria for the pi system (cyclic, planar, conjugated pi system with 4n+2 pi electrons). Both neutral heterocycles are aromatic. Perhaps most critically, given the options, in order to become non-aromatic when protonated, it means the basic site is a heteroatom lone pair that is part of the aromatic pi system and hence when the lone pair electrons are used in the bond to the proton, we change the aroma

Qu16:
Compounds that are anti-aromatic when deprotonated have non-aromatic conjugate bases. To be anti-aromatic, the conjugate acid would need to satisfy the first 3 criteria for aromaticity (i.e. it has a pi system that is cyclic, planar, fully conjugated) but contain 4n pi electrons = an even number of pi electron pairs. In this case, the conjugate base (i.e. the deprotonated form) which we can assume is planar due to the statement in the question, has 3 C=C and the lone pair (since it's a carbanion) hence 8 pi electrons (4n where n=2) or 4 pi electron pairs.

STARTING MATERIALS AND PRODUCTS:

If you are trying to find the product, then you should probably just work forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably the best way to go....
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu17:
Working backwards, in step 2, we have done a Clemmensen reduction that converts the C=O to a CH₂. The first step is an electrophilic aromatic substitution, specifically Friedel-Crafts acylation or alkylation (based on the AlCl3) but we see the aromatic unit, so what did we add ? Note that the reaction will be para to the Cl group (an o,p director). Based on the use of the Clemmensen and the nature of the alkyl group, we would need to acylate to avoid a rearrangement.

Qu18:
Working forwards, the first reaction is the dissolving metal reduction of the alkyne to give a trans-alkene followed by bromination to give a 1,2-dibromide via an anti addition. It's best to draw (or better still build a model) of product in the conformation in which it is formed in order to keep careful track of the stereochemistry, then convert that to the Fischer projection:

Qu19:
Working forwards, the first reaction is a hydride reduction of the nitrile to the primary amine. The nucleophilic amine will then react with the acid chloride to make the amide.

Qu20:
Working backwards.... The product is a cyclic system and enone (i.e. alkene ketone), which is potentially recognisable as the product of an aldol type condensation. This would be consistent with the reagents in step 3, a base and heat. "Reversing" the aldol reveals the keto-aldehyde which has been made by the ozonolysis of a cycloalkene (with a reductive work up)... counting C atoms will likely help here.

Qu21:
Working forwards, the reaction is the formation of an epoxide from the alkene, then reaction of the epoxide with a weak nucleophile (an alcohol) with an acid catalyst under SN1 like conditions (hence attack at the end that would gives the more stable cation character = benzylic position). It's best to draw (or better still build a model) of the product of the epoxide ring opening in the conformation in which it is formed first in order to manage the stereochemistry:

Qu22:
Working backwards, the product is a secondary alcohol made by the reaction of a Grignard reagent with an aldehyde (counting C atoms and thinking of the mechanism is the easiest way to make sure you get this right):

Qu23:
Working forwards, we have a disubstituted alkene being heated with a conjugated diene indicating a Diels-Alder reaction, since we have a good dienophile and step 1 just being "heat". Step 2 is a selective hydride reduction of the aldehyde carbonyl to the primary alcohol and step 3 the reduction of the alkene to the alkane. Make sure to count C atoms and use the "Diels-Alder template" to get the product. Since the dienophile was trans, the substituents on the product will be trans too.

Qu24:
Working forwards, the reaction is the formation of an epoxide from a halohydrin via an intramolecular SN2 (where the Nu (OH) needs to attack at 180 degrees to the Br (LG)). So, first, rotate the starting material into the reactive conformation with the -OH and -Br anti (180 degrees to each other). Now "make the reaction happen paying attention to the relative location of the alkyl groups. It's best to draw (or better still build a model) of the starting material in the reactive conformation to keep careful track of the stereochemistry:

Qu25:
The bromide needs to be eliminated to make the alkene to then form the 1,2-diol which is then reacted the the ketone to make the cyclic ketal:

Qu26:
Nitrate the aryl bromide (halogens direct o,p, get mainly para) then use diazonium chemistry to introduce the nitrile group:

Qu27:
Comparing the starting material and the product we've gained 1 C atom and there needs to be a rearrangement of the C skeleton... terminal alkyne can be converted into an aldehyde (via anti-Markovninkov hydration. Wittig reaction to add the C and make the alkene and then acid catalysed rearrangement:

Qu28:
Comparing the starting material and the substituted cyclohexene product and the bicyclic product suggest a Diels-Alder reaction but we've gained a C atom too... so enolate formation and alkylation, then the Diels-Alder then hydride reduction of the ketone to the secondary alcohol:

Qu29:
The keys issues are recognising that the cyclic acetal masks an aldehyde, we added a phenyl group, that the primary alcohol could be oxidised to give an aldehyde and an alkene could be formed by elimination to give the conjugated carbonyl, and perhaps that "H3O+ / heat" will i. complete the Grignard work up & ii. deprotect the acetal to reveal the aldehyde & iii. cause the elimination (heat) because of the stability of the conjugated product:

EXPLANATION OF PHENOMENA

Qu30:
Conjugated dienes and an electron withdrawing alkene (a dienophile) suggets a Diels-Alder reaction. Diels-Alder reactions require that the conjugated diene adopt the s-cis conformation to react, The s-cis conformation can be destabilised by cis-alkyl groups. Both X and Y are shown in the s-trans conformation. In structure X, the E-isomer, the cis-methyl group on the alkene when in the s-cis conformation is too close to the H atom of the terminal =CH₂ group so it destabilises the reactive s-cis conformation and this means less of X is in the reactive conformation, slowing the reaction with the dienophile, maleic anhydride.

Qu31:
The ester substituent is connected to the aromatic ring via the alkoxy O atom which will be moderately electron donating through resonance and directs o,p but sterics will hinder ortho attack.

Qu32:
Acids and bases undergo rapid proton transfer reactions. Given the pKas of the molecules involved (approx. acid pKa = 5, alcohol pKa = 15), the alkoxide will react with the carboxylic acid to give the carboxylate, but the aqueous acid work up will then just protonate the carboxylate to reform the carboxylic acid.

Qu33:
Both N atoms in the structure are sp2 hybridised. The lone pair on N1 is in an sp2 hybrid orbital and not involved in resonance or the aromatic pi system, while the N2 lone pair is in a p orbital, and in the aromatic pi system, and hence can be used to provide resonance stabilisation (shown below):.

Qu34:
Addition of bromine to alkenes and alkynes typically occurs in a stepwise manner via a cyclic bromonium ion which dictates that the addition of bomine occurs in an anti-fashion, giving the trans-1,2-dibromide.