I give it a try with your first question: If $A\in\mathbb{R}^{m\times n}$ then $\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m}\|A\|_\infty$ $\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n}\|A\|_1$
Also $\|AB\|_a\leq \|A\|_a\|B\|_a$
Note that $\|B_i\|_2<1$ implies that absolute value of the every element of matrix B is less than 1, $|B_{ij}|<1$, then $\|B\|_\infty (infinity norm of a matrix is the maximum absolute row sum of the matrix). And with the same argument $\|B\|_1.
Then using these properties we have $\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq \sqrt{nm} \|vA^T\|_2\|B\|_\infty\leq n\sqrt{nm} \|vA^T\|_2$ or $\|vA^TB\|_\infty\leq \sqrt{n}\|vA^TB\|_2\leq \sqrt{n} \|vA^T\|_2\|B\|_2\leq n \|vA^T\|_2\|B\|_1\leq nm \|vA^T\|_2$ (these are conservative bounds you can probably obtain a tighter bound)