DCFLs are not closed under intersection. (In fact, when we intersect two DCFLs we can get a non CFL also). Now, even for CFLs, checking if it is a DCFL is undecidable. So, A is undecidable.
Same as above, checking if a CFL is empty is decidable but after intersecting two DCFLs we can get a CSL which may not be a CFL. And checking whether a CSL is empty is undecidable.
$L(D_1)\subseteq L(E)$ is same as $L(D_1)\cap L(E)^c = \emptyset.$ $L(E)^c$ is regular since regular set is closed under complement and DCFLs (even CFLs) are closed under intersection with a regular language reducing our problem to checking if a DCFL is empty which is decidable.
$L(E)\subseteq L(D_1)$ is same as $L(D_1)^c\cap L(E) = \emptyset$ and like above this reduces to checking if a DCFL is empty which is decidable.
So, correct options: C;D
Reference: https://gatecse.in/grammar-decidable-and-undecidable-problems
