Hollow conductor containing charge: why is internal field cancelled outside and why are the field oustide the cavity zero inside the cavity?

Paraphrasing Griffith's "Second Uniqueness Theorem" (section 3.1.6, 4th ed.):

The electric field in a volume bounded by conductors is uniquely determined if the total charge on each conductor is given.

(1) Since the charge is only placed on B, this tells you that the field in the cavity is uniquely solved by the shape of the cavity and $Q$. In other words, the outside surface cannot contribute to the cavity: $\vec{E}_{A,ext}=0$. However, by the same reasoning I think it is wrong to say $\vec{E}_{A,int}=0$ in the cavity because the inner surface (and therefore $\sigma_{A,int}$) must contribute to uniquely solving the field.

(2) The outer surface will have $Q$ distributed on it regardless of the cavity's shape. So even though the outer conductor technically has no charge, just think of the volume outside A as bounded by A's exterior (with $Q$ on it) and infinity. Then apply the same uniqueness theorem to show the irrelevance of the cavity (in other words, $\vec{E}_{A,int}+\vec{E}_B=0$ outside).

Addressing your attempts:

(1) The rigorous explanation lies in the uniqueness theorems presented in Chapter 3, and their associated proofs.

(2) The final step is wrong: A vanishing integral does not imply a vanishing integrand in general. Besides, that's not a conclusion you want to come to. Otherwise it is good reasoning, and very much in the spirit of $\Phi$ as a conservative field (replacing $\vec{E}$ with $-\nabla \Phi$ makes the integrals very easy to compute moving between the surfaces of a single conductor).