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:
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.

Qu2:
Use the approx pKas of the acids: terminal alkyne CH = 25, ammonia NH = 35 and water = 15 or think about the bases that can be used to prepare an acetylide ion !
Remember that the hybridisation of the C (sp) makes it difficult to compare to the O and N systems where the electronegativity is a key factor.

Qu3:
Reaction is the Diels-Alder reaction and we are looking at the dienes reacting with the dienophile (benzoquinone). The reactivity increases in the Diels-Alder reaction with electron donating groups on the diene and the diene needs to be able to adopt an s-cis conformation. The simplest diene, 1,3-butadiene, has conformational flexibility and can readily rotate at room temperature into the reactive s-cis conformation. (2Z,4Z)-hexa-2,4-diene can rotate into the reactive conformation, but the terminal methyl groups would destablise that conformation, making that diene less reactive. 1,3-cyclopentadiene is locked in the reactive s-cis conformation and has electron donating groups attached at each end of the diene system. The molecules are drawn in the s-cis conformation below.

Qu4:
The hydroboration / oxidation of alkenes adds the HO and H across the C=C in an anti-Markovnikov fashion. Here we are asked about 3,3-dimethylbut-1-ene. The major product is the anti-Markovnikov product, followed by the Markovnikov product but there will be none of the rearranged product since the reaction does not proceed via an carbocation.

Qu5:
Alkenes undergo electrophilic addition. In the reactions with HX, the rate is controlled by the rate of carbocation formation. In this case, this means the acidity of the HX species and hence the stronger the acid the faster the carbocation is created.

HBr > HCl > CH₃CO₂H

Qu6:
This question relates directly to the sodium borohydride reactivity experiment from the laboratory where the reactivity aldehydes > ketones > esters.

Qu7:
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.

Qu8:
Carbocation stability.... Look at the C atom bearing the charge and what's attached to it. The vinyl carbocation is the least stable. Note that the Br can also stablise the +ve charge through resonance thus secondary and with Br attached is more stable than just secondary (reactions of vinyl halides with HX tells us this).

Qu9:
Conjugated dienes and more stable than isolated dienes (due to the interaction of the two pi systems), more highly substituted alkenes are more stable than less highly substituted alkenes:

Qu10:
All the bonds are C-C bonds. The factors involved are the hybridisation of the C atoms involved. C-C bonds involving two sp2 C atoms are shorter than two sp3 C atoms (due to higher s character).

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, we have a Simmons-Smith cyclopropanation of a C=C followed by the formation of a methyl ether by the reaction of an alcohol with the methyl chloride..

Qu12:
Working forwards, we have a halohydrin reacting with Na₂CO₃ (a weak base) to make an epoxide via an SN2 reaction. Recall that in an SN2, the Nu attacks at 180 degrees to the LG and therefore, the Cl on the opposite face to the -OH is going to be more reactive. The stereochemistry of the epoxide is defined by the stereochemistry of the initial C-O bond from the -OH group.

Qu13:
Working forwards, we have a terminal alkyne that we are treating with a strong base to make a nucleophilic acetylide which then undergoes an SN2 with an alkyl halide to make an internal alkyne.

Qu14:
Working forwards... The hydration of an alkyne initially gives an enol which tautomerises to give a ketone. Since it is an unsymmetrical alkyne with the O adding nearer the less hindered end, we should use a hydroboration using the more hindered reagent 9-BBN for best results.

Qu15:
Working forwards, when heated with acid, the alcohol dehydrates to give the more stable alkene (Zaitsev product). This is followed by alkene ozonolysis with a reductive work-up (we can tell the work-up is reductive because we have an aldehyde).

Qu16:
Working backwards...The structure of the product and the reaction conditions suggest a intramolecular Diels-Alder reaction. Pushing the arrows from the alkene in the product reveal the required starting material.

Qu17:
Working backwards... the product is an ether which could be made via a Williamson ether synthesis of an alcohol and an alkyl halide. If we are making the alcohol, it would need to be propan-1-ol which would require an anti-Markovnikov addition to an alkene.

Qu18:
Working forwards... Strong acid protonates the alkene and after a carbocation rearrangement (a 1,2-alkyl shift) to give the more stable tertiary carbocation, loss of a proton can give the alkene (Zaitsev product).

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:
The first step is halohydration of an alkene via an anti-addition and putting the HO- group at the tertiary position. Treatment with base converts the 1,2-halohydrin to an epoxide. The epoxide is then opened with methanol (weak Nu) with an acid catalyst via an SN1 like pathway to put the OMe group at the more substituted position and with trans-stereochemistry since the Nu attacks on the opposite face to the O of the epoxide.

Qu20:
The first step is a dissolving metal reduction of an alkyne, an anti addition, giving the trans-alkene. The trans-alkene is then reacted with osmium tetraoxide to give a 1,2-diol via a syn addition.

Qu21:
When reacted with a strong, bulky base, we get elimination of the alkyl bromide to give the anti-Zaitsev product, the less highly substituted alkene. Hydroboration / oxidation which gives the anti-Markovnikov alcohol (the primary alcohol).

Qu22:
The catalytic hydrogenation is a syn addition, giving the cis-alkene. This is followed by halohydration of an alkene via an anti-addition and putting the HO- group at the benzylic position (more stable +ve charge). Remember how to decode Newman projections. First identify the 1,2-halohydrins in the choices with the -OH in the benzylic position. Then rotate such that the -OH and -Cl are anti to reveal the required cis-alkene stereochemistry.

Qu23:
The structure of the product suggests a Diels-Alder reaction using cyclopenta-1,3-diene. But an alkyne dieneophile would give an isolated diene product. The trans-stereochemistry of the methyl groups in the product can only created by using a trans-alkene as the dienophile and therefore requires a dissolving metal reduction of an alkyne, an anti addition, giving the trans-alkene before the Diels-Alder reaction.

Qu24:
Reduction of an alkyne to an alkene and addition of Br2 with give a 1,2-dibromide via an anti addition. In order to get the required stereochemistry, redrawing the product with the Br atoms anti shows that you need a cis-alkene and therefore need to use a catalytic hydrogenation using Lindlar's catalyst to get that.

Qu25:
In order to form an epoxide from a 1,2-halohydrin, the -OH nucleophile needs to be at 180 degrees to the -Br LG. Therefore, we need to rotate the molecule into the reactive conformation about the middle C-C. Since the epoxides are shown with the O atom up, it is best to rotate the RHS with the Br atom. It is also a good idea to put the reacting groups (the -OH and -Br in the plane of the page). Hence the product has the alkyl group and the ester trans and the enantiomer shown.

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:
Hydration of alkenes with aq. sulfuric acid is controlled by the rate of carbocation formation where the most stable carbocation forms fastest. Here, the most stable cation that can form is tertiary and allylic.

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 and (5) alkynes are less stable than dienes (because the second pi bond in an alkyne is weaker than the pi bond in an alkene). Therefore the least stable isomer is the terminal, unbranched alkyne.

Qu30:

Qu31:
Tautomers (remember alkyne hydration ?) of ketones are the enols.

Qu32:
The addition of HCl to an alkene is a stepwise process with H+ adding to form a carbocation intermediate:

Qu33:
Organic E / Z nomenclature is based on the Cahn-Ingold-Prelog priority rules of the groups attached to the C=C.

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