Solution:

a. Since $B$ is given as a multiple of $A$, every prime that divides $A$ also divides $B$.

Conversely, suppose $p$ is a prime that divides $B$. Since $B = 2^{k} \cdot A$, either $p$ divides $2^{k}$ or $p$ divides $A$. If $p$ divides $2^{k}$, then $p = 2$. But then $p$ divides $A$ anyway, because $A = 2\left(2^{k-1} - 1\right)$. This shows that every prime that divides $B$ also divides $A$.

Since every prime that divides $A$ also divides $B$, and vice versa, $A$ and $B$ have the same set of distinct prime factors.

b. Observe that $A+1 = 2^{k} - 1$ and $B+1 = 2^{k}\left(2^{k} - 2\right) + 1 = 2^{2k} - 2 \cdot 2^{k} + 1 = \left(2^{k} - 1\right)^{2} = (A+1)^{2}$.

Since $B+1 = (A+1)^{2}$, every prime that divides $A+1$ divides $B+1$ and vice versa.

Therefore, $A+1$ and $B+1$ have the same set of distinct prime factors.