Can bridged and bridgehead positions be chlorinated using free radical mechanism?
The "bridged position" is just another ring, when viewed differently.

Suppose you look at the molecule a little differently (rotate it around a bit), then you could have a five-membered main ring, with a two membered "bridge", as opposed to what we have now (6 membered main ring with 1 membered bridge).

It's definitely possible to have a free radical halogen attack on that position, and with two stereoisomers at the positions which you have identified, it becomes total 3 monochlorinated derivatives.


(in case you're wondering how to check if the two stereoisomers are not identical, you could try to superimpose them, or go from one form to the other without breaking any bonds. In that case, you'll find that they're indeed different. You could even check for a plane/center of symmetry)

--EDIT--

My answer is wrong, as @Abcd and @AkshatJoshi pointed out, because I counted stereoisomers along with the required structural isomers.

I read about attacks at Bridgehead positions and there's NO place that mentions the existence of Bridgehead free radicals. Thus, there will only be 2 structural isomers, but 3 isomers overall.

So, I believe the answer ought to be 2.