Which of the following changes would make the radical chlorination of an alkane LESS selective (product distribution closer to the statistical one)?

Selectivity arises from small differences in activation energy, $k_1/k_2=e^{-\Delta \Delta E_a/RT}$. The exponent shrinks as T rises, so the ratio approaches 1 and the product ratio approaches the statistical one. Lower T and $\mathrm{Br}{}_2$ both raise selectivity, and a substrate with only one kind of H has nothing to be selective about.