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.)
In a previous, incorrect, version of this answer, I had claimed that the tangent map at the identity was surjective. This is of course false. This was used only to arrive at an inequality: $$ \dim\mathrm{Lie}(G) \leq \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B)~. $$
In fact, this inequality follows from Sard's theorem and the fact that the map $A \times B \to G$ is surjective, for if $\dim G > \dim A + \dim B$ then the image of the map $A \times B \to G$ would have measure zero.
Having established the above inequality, we now establish the reverse inequality. To see this simply notice that if $\mathrm{Lie}(A) \cap \mathrm{Lie}(B) = 0$, then the fact that $\mathrm{Lie}(A)$ and $\mathrm{Lie}(B)$ are subspaces of $\mathrm{Lie}(G)$ implies $$ \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B) \leq \dim\mathrm{Lie}(G)~. $$ So the answers to Q1 and Q2 are both true (as vector spaces not as Lie algebras).