Answer
$a.\displaystyle \quad\frac{\pi}{3}$
$b.\displaystyle \quad\frac{\pi}{3}$
Work Step by Step
$ a.\quad$
$\tan^{-1}x=y \ \Leftrightarrow\ \tan y=x$ and $y\in(- \displaystyle \frac{\pi}{2},\frac{\pi}{2})$
In $(- \displaystyle \frac{\pi}{2},\frac{\pi}{2}),\ y=\displaystyle \frac{\pi}{3}$ is such that $\displaystyle \tan\frac{\pi}{3}=\sqrt{3}$, so
$\displaystyle \tan^{-1}\sqrt{3}=\frac{\pi}{3}$
$ b.\quad$
$y=\sec^{-1}x\ ,\ (|x|\geq 1) \ \Leftrightarrow\ \sec y=x\ $ and $\ y\in[0, \pi/2) \cup[\pi, 3\pi/2)$
$y=\displaystyle \frac{\pi}{3}\in[0, \pi/2)$ and $\displaystyle \sec\frac{\pi}{3}=2$, so
$\displaystyle \sec^{-1}2=\frac{\pi}{3}$
