For D:-
if A is not CFL and B is CFL then A∪B will not be a CFL.
A= $a^nb^nc^n$:- Its not CFL
B= Sigma*
A union B=Sigma* which is regular hence CFL
So it can be CFL
For C:-
A= $a^nb^nc^n$ :- Its not CFL
B= phi
A intersection B is phi,which is regular hence CFL,
These are just counter examples to prove c and d are false.
A and b are true