(i) «$\mathrm{S_N2}$, one step»

curly arrow from the lone pair / negative charge on the O of $\ce{HO-}$ to the carbon atom bonded to Cl ✔

curly arrow from the C–Cl bond to the Cl atom ✔

transition state showing the partial bonds «HO···C···Cl» AND the negative charge «in square brackets» ✔

products: $\ce{Cl-}$ AND $\ce{CH3CH(OH)CH(CH3)2}$

(ii) the C–Br bond is weaker than the C–Cl bond «285 versus 324 $\mathrm{kJ\,mol^{-1}}$» ✔

«because» Br is larger / the C–Br bond is longer

OR

$\ce{Br-}$ is a better leaving group «than $\ce{Cl-}$»

OR

the activation energy is lower ✔

(iii) Homolytic fission: each atom keeps one of the bonding electrons / «neutral» radicals are formed ✔

Heterolytic fission: both bonding electrons go to one atom «Cl» / ions are formed ✔

(i) [1] each: nucleophilic attack arrow from HO-, arrow for C-Cl bond breaking, transition state with partial bonds and negative charge (curly arrows may be drawn in the transition state instead), products Cl- and 3-methylbutan-2-ol. Award [3 max] for a correct SN1 mechanism. Do not penalize HO and Cl not drawn at 180°. (ii) M1: C-Br weaker than C-Cl; M2: reason (larger Br / better leaving group / lower Ea). (iii) [1] each; award [1 max] if the two descriptions are reversed.