Yes. You can compute $RHom_A(G,X)$ as Hom-complex from a projective resolution $P^\cdot$ of $G$ to $X$. You have a natural map $Hom_A(G,X) \to RHom_A(G,X)$ induced by thinking of a map $G\to X_k$ as a map of the projective resolution. You know this map is an quasi-isomorphism (apply the spectral sequence to the double complex of Homs, taking homology with respect to the complex $P^\cdot$ first; the next page of the sequence is $Hom_A(G,X)$), so you just need to show that it's a B-module map. But that's clear from the construction.

Yes. You can compute $RHom_A(G,X)$ as Hom-complex from a projective resolution $P^\bullet$ of $G$ to $X$. You have a natural map $Hom_A(G,X) \to RHom_A(G,X)$ induced by thinking of a map $G\to X_k$ as a map of the projective resolution $P^\bullet\to X$. You know this map is an quasi-isomorphism (apply the spectral sequence to the double complex of Homs, taking homology with respect to the complex $P^\cdot$ first; the next page of the sequence is $Hom_A(G,X)$ by your observation on higher Exts), so you just need to show that it's a B-module map. But that's clear from the construction.

Yes. You can choose an injective resolution of $X$ as an $A\otimes B^{op}$-module and we can use it to compute $RHom_A(G,X)$ as a $B^{op}$-module. This is still an injective resolution of $X$ as an $A$-module as well, so its higher homology vanishes.

Yes. You can compute $RHom_A(G,X)$ as Hom-complex from a projective resolution $P^\cdot$ of $G$ to $X$. You have a natural map $Hom_A(G,X) \to RHom_A(G,X)$ induced by thinking of a map $G\to X_k$ as a map of the projective resolution. You know this map is an quasi-isomorphism (apply the spectral sequence to the double complex of Homs, taking homology with respect to the complex $P^\cdot$ first; the next page of the sequence is $Hom_A(G,X)$), so you just need to show that it's a B-module map. But that's clear from the construction.