Here is an post-mortem analysis / "how to" for the MT. 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:
Alkenes undergo electrophilic addition. In the reactions with aq. H₂SO₄, the rate is controlled by the rate of carbocation formation. In this case, this means the more stable the carbocation, the faster the reaction. The conjugated alkene will react the faster (via a resonance secondary carbocation) than the simple alkene. The alkyne is the least reactive since they would require the much less favourable termolecular pathway to avoid the formation of the very unfavourable vinyl carbocations. The ease of carbocation formation dictates the rate of reaction. In the case of alkynes, the vinyl C+ would be too unstable and a slower termolecular mechanism occurs, so the alkyne is the slowest.

Qu2:
Use the approx pKas of the acids: there is an internal alkyne, so the CH is likely about 45, a terminal alkyne = 25 and a amine NH = 35. The electronegativity of the atom the acidic H is attached to is a key factor (CH va NH) and the atom hybridisation. Remember the question wants in acidity order (not pKa order), the most acidic H in each structure is shown in bold.

Qu3:
Carbocation stability.... Look at the C atom bearing the charge and what's attached to it. We have a secondary benzylic carbocation, a simple tertiary carbocation and a phenyl carbocation. The simple secondary benzylic carbocation is the most stable (remember the extra resonance stabilisation). Then the tertiary carbocation is more stable than the phenyl carbocation.

Qu4:
The relative stability of diene systems.... in this case we are looking at dienes , their geometry and the degree of alkene substitution. Conjugated dienes are more stable than isolated dienes, and then cumulated dienes).

Qu5:
Identify the chirality centers (*) each of which can be R or S so here we have 3 configurations (one is meso), 2 and 1 (not chiral). Remember that the maximum number of configurational isomers is 2ⁿ where "n" is the number of stereocenters.

Qu6:
The question is about the hydrations of alkynes. First, draw the named product. Simple hydration of a terminal alkyne tends to give good yields of the methyl ketone (Markovnikov's rule). Simple internal alkynes on the other hand tend to give a more or less mixture of regioisomers. Hydroboration / oxidation of alkynes in this case a terminal alkyne will tend to give an aldehyde (which is the anti-Markovnikov product).

Qu7:
How many H atoms are there on the C atoms that are adjacent to the C=O group ? (alpha H are highlighted in blue below)

Qu8:
All about specific rotation. Looking at the three structures, we need to use the Cahn-Ingold-Prelog rules to assign the configurations. One is (R,R), one is (S,S) and the other is (R,S), a meso compound and therefore is optically inactive and has by definition a specific rotation = zero. Given the data in the question, the (S,S) has specific rotation of +13.2 degrees.

Qu9:
Reaction is the Diels-Alder reaction and we are looking at the dienophiles reacting with the diene (1,3-cyclopentadiene).  The reactivity increases in the Diels-Alder reaction with electron withdrawing groups on the dienophile. The carbonyl group (such as in an ester) is electron withdrawing through resonance (and two are better than one) while alkoxy groups (as in ethers) are electron donating through resonance.

Qu10:
The reaction is electrophilic addition of HBr to a conjugated diene via a carbocation under thermodynamic control (higher T). Protonation of the diene gives the most stable tertiary allylic carbocation. Under thermodynamic control, the major product is the more stable more highly substituted alkene formed by the reaction of the bromide ion nucleophile with the resonance contributor to the carbocation that had formed (i.e. 1,4- or conjugation addition). The second product is the direct addition product from the reaction of the bromide ion with the tertiary allylic carbocation.

STARTING MATERIALS, REAGENTS 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.

Qu11:
Working forwards : the unsymmetrical epoxide has undergone ring opening. Since the epoxide "O" has become the alcohol group, then the nucleophile "MeO" attacked the more hindered end, then this must have used acidic SN1 like conditions, CH₃OH / H+

Qu12:
Working forwards : the acid catalysed dehydration of an alcohol to give the more highly substituted Zaitsev alkene : conc acid and heat.

Qu13:
Working backwards : the product is a methyl ketone that looks to have been formed by the hydration of a terminal alkyne which in turn was formed by double elimination ( - 2 HBr to form 2 pi bonds)... count C atoms

Qu14:
Working forwards : the peracid converts the alkene to the epoxide. This undergoes ring opening to make the 1,2-diol then double dehydration will give the more stable conjugated diene.

Qu15:
Working forwards : a C3 alkene to a C5 secondary alcohol so we need to make a C-C bond. Ring opening an epoxide with a nucleophilic acetylide would do this and then we need to reduce the alkyne to the alkane chain.

Qu16:
Working forwards : allylic radical substitution adds the Br, then E2 elimination to give the conjugated diene. Step 3 is thermodynamically controlled 1,4- addition.

Qu17:
Working forwards : terminal alkyne alkylation = deprotonation of a terminal alkyne to make a good nucleophile followed by an SN2 to yield the internal alkyne.

Qu18:
Working forwards : elimination of an alkyl halide (Zaitsev elimination) forms an alkene that then undergoes ozonolysis with an oxidative work-up to give the ketone and carboxylic acid (count C atoms).

REGIO- and STEREOCHEMISTRY:
How well do you know your reagents ? Look at what has actually happened in terms of the reaction functional group transformation and then first look for any regiochemical issues then finally the stereochemistry last (it's the hardest to sort out).  In cases where more than one product is formed in equal amounts (e.g. the enantiomers), then both must be selected for full marks, part marks are given when only one of the pair is selected.

Advice : in each case draw the starting material in the conformation in which is reacts or the product in the conformation in which it is initially formed using wedge-hash diagrams. It is a good idea to draw the materials in such a way that the new bonds are in the plane of the page.  Once you have drawn the materials it may also be good for you to use model kits for these questions too.  Once you have drawn the materials in this way, you may need to consider rotations around sigma bonds to make your answer match the options.  An alternative approach could be to assign configurations to your drawn answer to compare them with the options - this can be slow and prone to error.

Qu19:
Working backwards : since potassium permanganate reacts with alkenes to give 1,2-diols via a syn addition, the alkene will need to be cis.

Qu20:
Working forwards : the alcohol product is formed via a hydroboration / oxidation of an alkene (anti-Markovnikov, syn addition).

Qu21:
Working forwards : addition of excess HBr to an alkyne follows Markovnikov's rule but here one needs to pay attention to the stability of the cationic character of the steps so the d+ develops on the benzylic C where resonance can stabilise the developing charge.

Qu22:
Working forwards : the alkene is converted to a 1,2-halohydrin via an anti addition with the -OH group at the more substituted C (reaction occurs via a cyclic chloronium ion with H₂O then reacting as the nucleophile and attacking the more substituted C in an SN1 like fashion).

Qu23:
Working forwards : the alkene is converted to a 1,2-halohydrin using the hypohalous acid via an anti addition with the -OH group at the benzylic C (reaction occurs via a cyclic chloronium ion with H₂O then reacting as the nucleophile and attacking the more substituted C in an SN1 like fashion). The halohydrin reacting with Na₂CO₃ (a weak base) to make an epoxide via an SN2 reaction so we need to make sure we review the halohydrin in the reactive conformation with the O nucleophile undergoing a backside attack to the C-Cl bond so -OH and -Cl anti to each other...

Qu24:
Working forwards : the alkene undergoes catalytic reduction to the cis-alkene. In step 2, we have a Simmonds-Smith reaction to make the cyclopropane (syn addition, alkene stereochemistry same as cyclopropane stereochemistry, cis).

Qu25:
Working forwards :the alkyne undergoes dissolving metal reduction to the trans-alkene. Then in step 2, form the 1,2-dibromide via an anti addition which gives us the meso compound. Now given that the dibromides are shown in Fischer projections, we need to convert to a Fischer projection (remember to position the alkyl chain (vertical) and back into the page with all the horizontal bonds coming out towards you) where in a meso compound the two -Br will need to be on the same side.

PI SYSTEMS:
All about knowing properties of pi systems such as resonance, structure, bonding, shapes, and terminologies etc.

Qu26:
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact (parallel orbitals, not perpendicular). The image below has highlighted the p orbitals (blue circles for top down view of vertical p orbitals.

Qu27:
Resonance contributors are derived via the delocalisation of the pi electrons across the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures.

Qu28:
Alkenes undergo electrophilic addition. In the reactions with HCl, the rate is controlled by the rate of carbocation formation with the most stable carbocation forming the fastest. The most stable carbocation will be the tertiary carbocation.

Qu29:
The key issues are (1) conjugated systems are more stable than isolated systems, (2) alkyl groups on pi bonds stabilise them, (3) trans isomer are more stable than cis isomers (steric effects), (4) s-trans is typically more stable than s-cis. Therefore the least stable isomer is the s-cis-cis,cis-conjugated diene.

Qu30:

Qu31:

All the bonds are CC bonds. The factors involved are (1) the hybridisation of the C atoms involved and (2) the bond order. All the bonds to consider here are single bonds with the shortest being the C-C Bonds involving two sp2 C atoms.

Qu32:
The addition of BH₃ to an alkene is a concerted process with B atom adding to the less hindered end :

Qu33:
The reaction is a catalytic hydrogenation. The reaction is fastest for the weakest pi bond and that is found in an alkyne... so the alkyne is the most reactive. In contrast, the C=C in arenes are less reactive than those in alkenes (alkenes can be reduced in the presence of arenes).

Qu34:
Resonance contributors are derived via the delocalisation of the pi electrons across the pi system and can be derived by pushing curly arrows. Use the rules for recognising resonance structures.