Textbook: Loudon 5th Ed. (2009)

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

To undergo a Diels-Alder reaction, a dienophile must be in s-cis conformation. 

The bicylic compound below is locked into s-trans conformation; it can never rotate into s-cis conformation and so can't undergo a Diels-Alder reaction.

Of the other two compounds, the middle compound most easily rotates into s-cis conformation, and so will undergo a Diels-Alder reaction the fastest. The compound on the right has steric strain when in s-cis conformation, and so won't do Diels-Alder as easily.

Chair forms can be difficult to draw, so it's a good diea to draw the "overhead view" first.

This is an S_N1 reaction, but with a twist.

The top reaction is a regular S_N1 reaction; the leaving group (Br^-) leaves and the nucleophile (CH₃OH) attacks the carbocation.

The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack. 

So instead of just one S_N1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.

There are separate explanations for SN1 and SN2.

S_N1: The S_N1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of S_N1 reaction.

S_N2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all. 

The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an S_N2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of S_N2 reaction.

There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).

Also, the carbons on the double bond of allyl and benzyl compounds are sp² hybridized, and so are more electronegative than sp³ carbons. So the sp² carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing S_N2 reactivity.

One of the rules for aromaticity is that all atoms shoudl be sp² hybridized. But the nitrogen in pyrrole is sp³ hybridized, so how is it still aromatic? Because in 4/5 of its resonance forms the nitrogen is sp² hybridized; the real picture of pyrrole looks more like the structure on the left (dashed circle) than any individual resonance form.