Mechanism of nucleophilic substitution

The substrate, 1-bromo-2-methylpropane, is a primary alkyl halide.
An $S_N1$ reaction would need a primary carbocation, which is very unstable. In an $S_N1$ process this cation would rearrange (hydride shift) to the tertiary cation and give a rearranged product.
In both reactions the product is $(CH_3)_2CH-CH_2OC_2H_5$, in which the carbon skeleton is unchanged; no rearrangement has occurred.
So in both cases the nucleophile attacks the carbon bearing bromine directly, from the back side, in a single step: the $S_N2$ mechanism.
In (ii) the strong nucleophile $C_2H_5O^-$ clearly favours $S_N2$; in (i) the weaker nucleophile $C_2H_5OH$ reacts more slowly but, as the unrearranged product shows, also by $S_N2$.
Hence the mechanisms are $S_N2$ and $S_N2$.

In an $S_N1$ reaction the rate-determining step is the ionisation of the C–Br bond to form a carbocation; the more stable the carbocation, the faster the reaction.
$C_6H_5CH_2Br$ gives $C_6H_5CH_2^+$: a primary benzylic cation stabilised by one phenyl group.
$C_6H_5CH(CH_3)Br$ gives $C_6H_5\overset{+}{C}H(CH_3)$: a secondary benzylic cation (one phenyl + one methyl).
$C_6H_5CH(C_6H_5)Br$ gives $(C_6H_5)_2CH^+$: a secondary cation stabilised by resonance with two phenyl groups.
$C_6H_5C(CH_3)(C_6H_5)Br$ gives $(C_6H_5)_2\overset{+}{C}(CH_3)$: a tertiary cation stabilised by resonance with two phenyl groups and by the +I and hyperconjugation effects of the methyl group. It is the most stable.
Hence $C_6H_5C(CH_3)(C_6H_5)Br$ is the most reactive towards $S_N1$.