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.