Let $i_A : A \to G$ and $i_B : B \to G$ be the embeddings and let $\mu :  G \times G \to G$ be the group multiplication.  Then the composition
$$
A \times B \stackrel{i_A\times i_B}{\longrightarrow} G \times G \stackrel{\mu}{\longrightarrow} G
$$
is a smooth map which is surjective and sends the identity to the identity.  (It is not, however, a group homomorphism.)  Its tangent map at the identity is a surjective linear map $$
\mathrm{Lie}(A) \oplus \mathrm{Lie}(B) \to \mathrm{Lie}(G)~,
$$
hence
$$
\dim\mathrm{Lie}(G) \leq \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B)~.
$$
But then if $\mathrm{Lie}(A) \cap \mathrm{Lie}(B) = 0$, then
$$
\dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B) \leq \dim\mathrm{Lie}(G)~.
$$
So the answers to Q1 and Q2 are both true (as vector spaces not as Lie algebras).

I am not sure what references to point to.  It seems this is just basic Lie groups as manifolds stuff.