(i) Between $\mathrm{CH}_3\mathrm{Br}$ and $\mathrm{CH}_3\mathrm{I}$, $\mathrm{CH}_3\mathrm{I}$ will react faster via the $\mathrm{S}_\mathrm{N}2$ mechanism. In $\mathrm{S}_\mathrm{N}2$ mechanism, C – X bond breaks and the faster it breaks faster is the reaction. $\mathrm{I}^{-}$ is a better leaving group. Owing to its large size, the C – I bond breaks faster than the C – Br bond and reaction proceeds further at a greater rate. (ii) The order of reactivity in an $\mathrm{S}_\mathrm{N}2$ reaction depends on minimal steric hindrance around the carbon involved in the C – X bond. Lesser the steric hindrance felt by the incoming nucleophile,more reactive will be the alkyl halide towards $\mathrm{S}_\mathrm{N}2$ reaction. Based on this, $\mathrm{CH}_3\mathrm{Cl}$ will react faster than $(\mathrm{CH}_3)_3\mathrm{CCl}$.