I don’t know how useful it is, but here’s a slightly less ugly answer to the first question.
Let $S=\{a_1,\dots,a_k\}$ with $a_1<\dots. A set $B=\{a_i,a_{i+1},\dots,a_{i+m}\}$ of consecutive members of $S$ is a block of $S$ iff $a_{i+j}=a_i+j$ for $0\le j\le m$ and $a_i-1,a_{i+m}+1\notin S$. Let $M$ be the multiset whose members are the sizes of the blocks of $S$. Then $|\{\pi\in_n:\operatorname{desc}(\pi)\supseteq S\}|= \frac{n!}{\prod\limits_{m\in M}(m+1)!}.\tag{1}$
To see this, let $B=\{a_i,a_{i+1},\dots,a_{i+m}\}$ be a block of $S$ of size $m+1$. The $m+2=|B|+1$ numbers $a_i,\dots,a_{i+m+1}$ must appear in descending order in $\pi$, a condition that is satisfied by only $\frac1{(m+2)!}$ of the $n!$ permutations of $[n]$. The corresponding conditions for any other blocks of $S$ are independent, and $(1)$ follows immediately.
More laboriously, one can observe that if $B$ is a block of $S$ of size $m$, there are $\binom{n}{m+1}$ ways to choose the members of the block and the immediately following element of $\pi$ and no freedom in how they are ordered. The remaining $n-\sum\limits_{m\in M}(m+1)$ elements of $\pi$ can be ordered arbitrarily, so we end up with $\left(n-\sum_{m\in M}(m+1)\right)!\prod_{m\in M}\binom{n}{m+1}=\frac{n!}{\prod\limits_{m\in M}(m+1)!}$ permutations whose descents include $S$.