Yes, this video is a good deterrent for anyone who is ever thinking of working with F2.

Draw the molecule and look for primary, secondary, tertiary C-H bonds. Ask: is bromination selective for any of these positions?

Chlorine radicals are less selective than bromine radicals. https://www.masterorganicchemistry.com/2013/10/31/selectivity-in-free-radical-reactions-bromine-vs-chlorine/

Free-radical reactions require a radical “initiator” that breaks bonds homolytically. This can be either “light” (hv) or “heat” in the presence of various radical “initiators” with weak bonds, such as “peroxides” (RO-OR) or other species like AIBN. https://en.wikipedia.org/wiki/Azobisisobutyronitrile

Free radical substitution will end up on the benzylic carbon. See this post: https://www.masterorganicchemistry.com/2018/06/13/reactions-on-the-benzylic-carbon-bromination-and-oxidation/

That’s a question you can certainly figure out for yourself! Start by drawing out butane with all its hydrogens and carbons, and replacing two hydrogens with chlorines on the leftmost molecule. That’s 1,1-dichlorobutane. Now systematically keep trying to make new dichlorobutanes until you can’t make any new ones!

You will get a mixture of both. The reaction is not selective!

Free radicals can usually invert their stereochemistry, so any stereochemical information is usually lost. They are somewhat like carbocations in this regard.