Answer
$a.\displaystyle \quad-\frac{\pi}{4}$
$b.\displaystyle \quad\frac{\sqrt{3}}{2}$
Work Step by Step
$ a.\quad$
$\displaystyle \tan\frac{3\pi}{4}=-1$
$\tan^{-1}(-1)=y \Leftrightarrow \tan y=-1$ and $y\in(- \displaystyle \frac{\pi}{2},\frac{\pi}{2})$
In $(- \displaystyle \frac{\pi}{2},\frac{\pi}{2}),\ y=-\displaystyle \frac{\pi}{4}$ is such that $\displaystyle \tan(-\frac{\pi}{4})=-1$, so
$\displaystyle \tan^{-1}\tan(\frac{3\pi}{4})=\tan^{-1}(-1)=-\frac{\pi}{4}$
$ b.\quad$
$\arcsin x=\sin^{-1}0.5=y \ \Leftrightarrow\ \sin y=0.5\ $ and $\ y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$
In $[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}],\ y=\displaystyle \frac{\pi}{6}$ is such that $\sin y=0.5$, so
$\displaystyle \cos(\arcsin\frac{1}{2})=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$