Let us see each of the options one by one..
A) is decidable as we have various membership algos for CFL class of languages like CYK algo , LL(k) and LR(k) parsing algos etc..In fact the upper bound to determine if a string belongs to a CFL is given by O(n3) in worst case scenario by CYK algorithm and in some cases we have best case as O(n) as is the case of S - grammar..
B) Let us assume that a CFL is regular..For that we have to establish the fact that there exists some regular language L2 for CFL L1 ..The equivalence can be written mathematically as :
L1 ⊕ L2 = Φ = L3 (say)
==> (L1 - L2) U (L2 - L1) = Φ
Thus if we are able to establish L3 to be empty is decidable, then our claim will be true..
Now we concentrate on the 2nd term we are performing L2 - L1 which can be written as : L2 ⋂ L1'
Now L1 being a CFL we know is not closed under complementation..So we go up the Chomsky hierarchy for the sake of generality and now hence we push L1 to CSL level..
Now L2 is regular and L1' is a CSL for sure..So the 2nd term will also be a CSL and the first term (L1 - L2) will be a CFL..
But to do union , we push the 1st term up to CSL level , hence L1 - L2 be CSL..Now union of 2 CSL is a CSL..
Hence the resultant language L3 is a CSL..
But we know emptyness problem for CSL is undecidable..
Hence our assumption was false..
Hence it is proved that the regularity of CFL is undecidable..
C) Now coming to C).This is a trivial problem as finite state automation is an specific case of halting turing machine with limited power..Hence it is decidable problem..
D) Type 0 language means recursively enumerable language..But all recursively enumerable languages are not recursive and hence cannot be necessarily modelled by Halting Turing Machine..So possibility is that for some cases of rejection of input , the Turing Machine will not halt..Thus making the membership property undecidable for RE languages..
Therefore the correct answers are B) and D)