Answer
continuous on $(-2, -\pi/2)\cup(-\pi/2, \pi/2)\cup(\pi/2,2)$
Work Step by Step
$f(x)=\tan x$ is continuous on its domain, $\displaystyle \{x|x\neq\frac{\pi}{2}+\pi\pi\}$
(Theorem 7$: $polynomials, rational functions, root functions, trigonometric functions are continuous on their domains)
$v(x)=4-x^{2}$, a polynomial is continuous everywhere (Th.5)
$u(x)=\sqrt{x}$, is continuous for $x \geq 0.$
$g(x)=u(v(x))$, a composite of continuous functions is continuous by Th.9.
The domain of $g(x)$ is
$4-x^{2}\geq 0,$
$4\geq x^{2}$
$x\in[-2,2]$
$B(x)=\displaystyle \frac{f(x)}{g(x)}$ has domain:
$ x\displaystyle \neq\frac{\pi}{2}+\pi\pi,$ and $ x\in(-2,2),$
(not defined for $\pm 2$ because of zero in the denominator,
and
the domain of f demands $\displaystyle \pm\frac{\pi}{2}$ also to be excluded.)
Domain= $(-2, -\pi/2)\cup(-\pi/2, \pi/2)\cup(\pi/2,2)$
On this domain B(x) is continuous by Th.4.5.
