This seems too easy for Q3, so I'm probably wrong. We word this with sets/posets. So think of the factors as sets of the primes. (so a prime is in the set of the factor if the prime divides it) So call the sets $s_1, s_2, \ldots, s_{2^k}$, and the corresponding factors are $f_1, f_2, \ldots, f_{2^k}$. Note that if $a \le f_i, f_j \le b$ and $s_i \subset s_k \subset s_j$, then $a \le f_k \le b$ (*) (This is trivial) Now cover the poset $s_1, s_2, \ldots, s_{2^k}$ with chains (specifically $\binom{k}{k/2}$ chains). This is standard and is pretty much Sperner's Theorem and Dilworth's Theorem. For each chain $C$ let $C_1, C_2$ be the sets of subsets that have even, respectively odd size. So for each chain, the difference $|C_1| - |C_2| \le 1$, because of (*). Now sum over all $\binom{k}{k/2}$ chains to get the result.

This seems too easy for Q3, so I'm probably wrong. We word this with sets/posets. So think of the factors as sets of the primes. (so a prime is in the set of the factor if the prime divides it) So call the sets $s_1, s_2, \ldots, s_{2^k}$, and the corresponding factors are $f_1, f_2, \ldots, f_{2^k}$. Note that if $a \le f_i, f_j \le b$ and $s_i \subset s_k \subset s_j$, then $a \le f_k \le b$ (*) (This is trivial) Now cover the poset $s_1, s_2, \ldots, s_{2^k}$ with chains (specifically $\binom{k}{k/2}$ chains). This is standard and is pretty much Sperner's Theorem and Dilworth's Theorem. For each chain $C$ let $C_1, C_2$ be the sets of subsets that have even, respectively odd size. So for each chain, the difference $|C_1| - |C_2| \le 1$, because of (*). Now sum over all $\binom{k}{k/2}$ chains to get the result.