Answer
$a.\displaystyle \quad\frac{5\pi}{6}$
$b.\displaystyle \quad\frac{\pi}{2}$
Work Step by Step
$ a.\quad$
$y=\cot^{-1}x,\ (x\in \mathbb{R}) \ \Leftrightarrow\ \cot y=x\ $ and $\ y\in(0, \pi)$
$y=\displaystyle \frac{\pi}{6}\in(0, \pi)$ is such that $\displaystyle \cot\frac{\pi}{6}=+\sqrt{3}$,
By symmetry, $\displaystyle \cot\frac{5\pi}{6}=-\sqrt{3} ,$
which means that $\quad\cot^{-1}(-\sqrt{3})$=$ \displaystyle \frac{5\pi}{6}$
$ b.\quad$
$\arcsin 1=\sin^{-1}1,$
$\sin^{-1}x=y \ \Leftrightarrow\ \sin y=x\ $ and $\ y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}].$
$y=\displaystyle \frac{\pi}{2}\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$ is such that $\displaystyle \sin\frac{\pi}{2}=1$, so
$\displaystyle \arcsin 1=\frac{\pi}{2}$