Reformulating the Given Conditions in Decidability Problems

I came across the following question:

Given two context-free languages $$L_1$$ and $$L_2$$ is it decidable whether $$L_1 - L_2 = \emptyset$$ ?

The problem $$ALL_{\text{CFG}}$$ that states:

Given a CFG $$G$$ then $$L(G) = Σ^*$$ is known to be undecidable.

So I thought to fix $$L_2$$ to be $$Σ^*$$ and allow $$L_1$$ to be any context-free language to prove that the problem stated is undecidable.

What concerns me the most is that the $$ALL_{\text{CFG}}$$ refers to a context-free grammar and not a context-free language.

One of the definitions of a context-language is the following:

$$L$$ is a context-free language $$\leftrightarrow$$ There exists a context-free grammar $$G$$ s.t $$L(G) = L$$ which translates to 2 things:

1. $$L$$ is a context-free language $$\rightarrow$$ There exists a context-free grammar $$G$$ s.t $$L(G) = L$$

2. Given a context-free grammar $$G$$ s.t $$L(G) = L$$ $$\rightarrow$$ $$L$$ is a context-free language

The only way I see to address my concern regarding $$ALL_{\text{CFG}}$$ is to add bullet $$(1)$$ to the proof and state that if the problem in question is decidable then $$L(G_1) = Σ^*$$ is decidable (which is not) where $$G_1$$ is the context-free grammar that generates $$L_1$$. Would this approach work?

I am seeing this kind of question for other kinds of languages as well.

Say that the problem in question was the following:

Given two recursive languages $$L_1$$ and $$L_2$$ is the $$X$$ problem decidable?

Will the same approach work if I manage to take the problem from the domain of recursive languages to the domain of Turing Machines with $$M_1$$ and $$M_2$$ where $$L(M_1) = L_1$$ and $$L(M_2) = L_2$$?

Given two context-free languages $$L_1$$ and $$L_2$$ is it decidable whether $$L_1 - L_2 = \emptyset$$ ?

It is understood the $$L_1$$ is given as $$L(G_1)$$ where $$G_1$$ is a given context-free grammar. $$L_2$$ is given as $$L(G_2)$$ where $$G_2$$ is a context-free grammar.

This convention is used for similar propositions/exercises unless specified otherwise.

For example, if we are asked "given two recursive languages $$L_1$$ and $$L_2$$ is the $$X$$ problem decidable?", then we can assume $$L_1$$ is given by $$L(M_1)$$ where $$M_1$$ is a decider and $$L_2$$ is given by $$L(M_2)$$ where $$M_2$$ is a decider.

Technically, "a context-free language $$L_1$$" can be for example given as $$L(S)$$ where $$S$$ is a context-sensitive grammar such that $$L(S)$$ is context-free or given as $$L(T)$$ where $$T$$ is a Turing machine such that $$L(T)$$ is context-free. However, these kinds of interpretations are not used anywhere during introductory study.

Here is an exercise.

Prove the following promise problem is undecidable.
Given a Turing machine $$T$$ that accepts a decidable language and a string $$w$$, determine whether $$T$$ accepts $$w$$.

Here we should understand that it is possible the given $$T$$ is not a decider., although $$T$$ accepts a decidable language.

This undecidability says there is no decider that given an arbitrary string $$w$$ and an arbitrary semi-decider $$S$$ that accepts a decidable language, can output whether $$S$$ accepts $$w$$. In other words, the halting problem is still undecidable even the Turing machine in the input is guaranteed (by an oracle) to accepts a decidable language each time (however, it is possible that the given Turing machine itself is not a decider).