Converting bromobenzene to 1‐bromo‐3‐methylbenzene

This is one of those pointless paper exercises that you would never do in the real world. I would not start from bromobenzene as $ce{Br}$ directs ortho/para, so this is going to take several steps.

You could make the phenyl Grignard or phenyllithium from bromobenzene quenching with DMF followed by aqueous acid workup to give benzaldehyde. Brominate this with $ce{BrCl}/ce{AlCl3}$ described here to give 3-bromobenzaldehyde then do a Wolff–Kishner reduction $(ce{H2NNH2}/ce{KOH})$ on the aldehyde to give 3-bromotoluene. I would not risk a Clemmensen $(ce{Zn}/ce{Hg}$ & conc. $ce{HCl})$ here as I think there is a danger of reducing the $ce{Ar-Br}$ to $ce{Ar-H}.$