Answer
a)$-2$
b)$6$
Work Step by Step
a) I know that $\tan$ is an odd function, which means $f(-\theta)=-f(\theta).$
Therefore $f(-a)=\tan{(-a)}=-\tan{a}=-2$.
b) I know that $\tan$ has a period of $\pi$, hence $\tan{\theta}=\tan{(\theta+k\pi)}$ where $k$ is an integer.
Thus $f(a)+f(a+\pi)+f(a+2\pi)=\tan{(a)}+\tan{(a+\pi)}+\tan{(a+2\pi)}=\tan{(a)}+\tan{(a)}+\tan{(a)}=2+2+2=6$
