(a)
The $\mathrm{S_N1}$ path begins with the C to Br bond breaking on its own to make a carbocation, so the isomer that forms the steadiest carbocation reacts fastest. Among the $\mathrm{C_4H_9Br}$ isomers, the one where bromine sits on a carbon joined to three other carbons is 2-bromo-2-methylpropane, $\mathrm{(CH_3)_3C\text{-}Br}$. When its bromide leaves, the leftover positive carbon is a tertiary carbocation, which is propped up by the electron push of three methyl groups. Being the most settled carbocation, it forms most readily, so this isomer is the most reactive towards $\mathrm{S_N1}$.
(b)
Sodium driven coupling (the Wurtz reaction) glues together two identical alkyl pieces, dropping the two bromine atoms as NaBr. To end up with 2,5-dimethylhexane, just cut that target down the middle into two equal halves. Each half is a $\mathrm{(CH_3)_2CH\text{-}CH_2\text{-}}$ unit, which is the isobutyl group. So the starting halide must carry bromine on that primary carbon, namely 1-bromo-2-methylpropane, $\mathrm{(CH_3)_2CHCH_2Br}$ (isobutyl bromide). Two of these join through sodium to rebuild the eight carbon skeleton:
\[ 2(CH_3)_2CHCH_2Br + 2Na \rightarrow (CH_3)_2CHCH_2CH_2CH(CH_3)_2 + 2NaBr \]
So (a) is tert-butyl bromide and (b) is isobutyl bromide.