1. L = { <G1, G2> | G1 & G2 are regular grammar and L(G1) ⊆ L(G2)}
True. Subset problem is decidable for regular grammars. You can use the method in (2) and (3) to prove the same.
2. L = { <G, R> | G is a CFG & R is a regular expression and L(G) ⊆ L(R)}
True. Subset can be rewritten as $L(G)\cap \overline{L(R)}=\phi$.
We know that Regular languages are closed under complement. And for a CFL, the intersection with Regular language is also closed. Now we know that $L(G)\cap \overline{L(R)}$ is a CFL. So create the CFG of that. $L(G) = \phi$ is decidable for CFG.
3. L = { <G, R> | G is a CFG & R is a regular expression and L(R) ⊆ L(G)}
False. We can use a similar approach of that of (2). Here $L(R) \cap \overline{L(G)}$. But the issue here is L(G) is not closed under complement. So we dont know what kind of language this is. Therfore undecideable.
So option(B) 1&2 only.
References:
(1) https://gatecse.in/grammar-decidable-and-undecidable-problems/
(2) https://gatecse.in/closure-property-of-language-families/
