How to determine the reactivity of a set of compounds that may undergo nucleophilic substitution?
In question 1), you are wrong about the tendency of leaving groups. While you are correct that the carbon-chlorine bond is more polarised towards chlorine, iodide is still a better leaving group. This is because it is much larger and can thus stabilise the resulting negative charge much better.

You can observe this in $\mathrm{S_N2}$ reactions, where iodides are much more reactive than bromides which are more reactive than chlorides (and fluorides are basically unreactive) even though the same polarisation argument could have applied.



In question 2), the stabilisation that alkyl groups give carbocations, or the electron density they donate into a phenylic ring actually comes from $\ce{C-H}$ bonds. Thus, the para-xylene derivative can stabilise a cation better since there are three $\ce{C-H}$ bonds it can use for stabilisation. No matter which way the methyl is oriented, it can stabilise. para-ethylbenzyl bromide can only be stabilised in two out of three orientations (when a $\ce{C-H}$ bond is orthogonal to the phenyl ring) and para-isopropylbenzyl bromide only in one. Thus, the rate of hydrolysis is accordingly.