D) L1 is CFL but L2 is not CFL.
L1=>
L1 can be accepted by a PDA and hence is CFL. But we need a NPDA for this as there is no deterministic way to identify where x ends and xR starts. So L1 is accepted by a NPDA and hence is NCFL.
L2=>
The set of strings in L2 are {aa,bb,aaaa,abab,baba,bbbb,aaaaaa...........}. We cannot accept these strings using an NFA. Now, even a PDA is not possible as once we store x on stack, it can only be read back in reverse order. Thus, we require 2 stacks to recognize L2. Now, L2 can be accepted by a TM in linear space and hence L2 is CSL.