equivalence classes of arch diagrams in bijection with permutations

Let me address Q2.

Paint all right endpoints in white, and all left ones in black. Number white points $1,2,\dots,n$ from left to right, and number the black points according to the numbers of white points joined to them.

All black points lying between two consecutive white ones can be permuted arbitrarily. So the only thing that matters for a black point is in which of the $n$ intervals (before 1, and between $i$ and $i+1$ for some $i$) it is situated. For the $i$th black point, there are $i$ possible options. Altogether, we get $1\cdot 2\cdots n=n!$ variants.