Answer
For $f+g, f-g, fg$: $D=\{x|-3 \leq x \leq 5\}$,
For $\frac{f}{g}$: $D_{\frac{f}{g}}=\{x|-3 \lt x \leq 5\}\}$,
Work Step by Step
$f(x)=\sqrt {25-x^2}$,
$D_f=\{x|-5\leq x\leq 5 \}$,
$g(x)=\sqrt {x+3}$,
$D_g=\{x|x \geq -3\}$,
thus,
$f+g=\sqrt {25-x^2} + \sqrt {x+3}$,
$D_{f+g}=\{x|-3 \leq x \leq 5\}$,
$f-g=\sqrt {25-x^2} - \sqrt {x+3}$,
$D_{f-g}=\{x|-3 \leq x \leq 5\}$,
$f \times g=(\sqrt {25-x^2})(\sqrt {x+3})=\sqrt {(25-x^2)(x+3)}$
$D_{f \times g}=\{x|-3 \leq x \leq 5\}$,
$\frac{f}{g}=\frac{\sqrt {25-x^2}}{\sqrt {x+3}}=\sqrt {\frac{25-x^2}{x+3}}$,
$D_{\frac{f}{g}}=\{x|-3 \lt x \leq 5\}\}$,
