Answer
For $f+g, f-g, fg$: $D=[-4,-1] \cup [1, 4]$,
For $\frac{f}{g}$: $D_{\frac{f}{g}}=[-4, -1) \cup (1, 4]$
Work Step by Step
$f(x)=\sqrt {16-x^2}$,
$16-x^2\geq 0$ for $x\in[-4,4]$
$D_f=[-4, 4]$,
$g(x)=\sqrt {x^2-1}$,
$x^2-1\geq 0$ for $x\in(-\infty, -1] \cup [1, \infty)$
$D_g=(-\infty, -1] \cup [1, \infty)$,
thus,
$f+g=\sqrt {16-x^2} + \sqrt {x^2-1}$,
$D_{f+g}=[-4,-1] \cup [1, 4]$,
$f-g=\sqrt {16-x^2} - \sqrt {x^2-1}$,
$D_{f-g}=[-4,-1] \cup [1, 4]$,
$f \times g=(\sqrt {16-x^2})(\sqrt {x^2-1})=\sqrt {(16-x^2)(x^2-1)}$
$D_{f\times g}=[-4,-1] \cup [1, 4]$,
$\frac{f}{g}=\frac{\sqrt {16-x^2}}{\sqrt {x^2-1}}=\sqrt {\frac{16-x^2}{x^2-1}}$,
$x^2-1=0\Rightarrow x=\pm 1$
$D_{\frac{f}{g}}=[-4, -1) \cup (1, 4]$.
