Fix a finite alphabet $\Sigma$ for the entire discussion.
There is a rather obvious proof that the difference of two context-free grammars $A$ and $B$ (compute the language $L(A) - L(B)$) is not computable by a reduction to the universality problem: to check if $A$ is universal check if $\Sigma^* - L(A)$ is empty.
I am interested if there is a more direct proof of this. Subtraction seems "hard" on its own. In particular, it stands to reason that the following restriction of the problem is also unsolvable:
For two CFGs $A$ and $B$ that are both infinite and co-infinite (complement of their language is infinite) compute $L(A) - L(B)$.
The above reduction to universality now doesn't work, since $A$ and $B$ are by definition not universal (or their complement would be the finite empty language).
I would assume that this is analogous to the problem of equality of two infinite and co-infinite grammars, which has a similar easy reduction to universality that does not work under the co-infiniteness restriction. In particular equality trivially reduces to subtraction.