(i) In the S${}_N$2 mechanism, the reactivity of halides for the same alkyl group increases in
the order. This happens because as the size increases, the halide ion becomes a better
leaving group.
R−F << R−Cl < R−Br < R−I

$C{H}_3$
$|$
$H_{3}C—C—Cl C{H}_3—Cl$
$|$
$C{H}_3$
Therefore, CH${}_3$I will react faster than CH${}_3$Br in S${}_N$2 reactions with OH−. (ii)
The S${}_N$2 mechanism involves the attack of the nucleophile at the atom bearing the leaving
group. But, in case of (CH${}_3$)${}_3$CCl, the attack of the nucleophile at the carbon atom is
hindered because of the presence of bulky substituents on that carbon atom bearing the
leaving group. On the other hand, there are no bulky substituents on the carbon atom
bearing the leaving group in CH${}_3$Cl. Hence, CH${}_3$Cl reacts faster than (CH${}_3$)${}_3$CCl in S${}_N$2
reaction with OH${}^{-}$.