I am giving it an attempt but I am pretty new to this topic..so anyone finding it wrong is requested to rectify me and give the correct explanation.

We know that cfls are not closed under intersection. So L(G1) ⋂ L(G2) may or may not be a cfl. But we can always say that L(G1) and L(G2) are recursively enumerable languages and so their intersection is also RE(as RE languages are closed under intersection).

Let L(G)= L(G1) ⋂ L(G2) where L(G) is RE. So now we can reduce the problem statement to "L(G)=Φ is decidable or not".

By Rice Theorem, every non trivial property of RE set is undecidable. For a property to be non trivial there should exist atleast 2 RE languages(hence 2 TMs),the property holding for one (T_{yes} being its TM) and not holding for the other(T_{no} being its TM).

T_{yes} is the TM whose language is Φ. Can T_{yes} exist? yes

T_{no} is the TM whose language is (a+b)*. Can T_{no} exist? yes

So we can say that this property is non trivial. And hence we can conclude that whether a RE language is Φ is an undecidable problem.

=> L(G)=Φ is undecidable

=> L(G1) ⋂ L(G2) =Φ is undecidable