Answer
$(f+g)(x)=\sqrt{25-x^{2}}+\sqrt{x+3}$ , domain: $[-3,5]$
$(f-g)(x)=\sqrt{25-x^{2}}-\sqrt{x+3}$ , domain: $[-3,5]$
$(fg)(x)=\sqrt{25-x^{2}}\cdot\sqrt{x+3}$ , domain: $[-3,5]$
$(\displaystyle \frac{f}{g})(x)=\frac{\sqrt{25-x^{2}}}{\sqrt{x+3}}$ , domain: $(-3,5] $
Work Step by Step
The domain of f is $\{x|-5\leq x\leq 5\}=[-5,5]$, because
f is defined only for x that yield a nonnegative number under the square root.
The domain of g is $[-3,+\infty )$, for the same reason.
Their intersection is $[-3,5]$
$(f+g)(x)=\sqrt{25-x^{2}}+\sqrt{x+3}$ , domain: $[-3,5]$
$(f-g)(x)=\sqrt{25-x^{2}}-\sqrt{x+3}$ , domain: $[-3,5]$
$(fg)(x)=\sqrt{25-x^{2}}\cdot\sqrt{x+3}$ , domain: $[-3,5]$
$(\displaystyle \frac{f}{g})(x)=\frac{\sqrt{25-x^{2}}}{\sqrt{x+3}}$ , domain: $(-3,5]$ , because $g(-3)=0$, so $-3$ is excluded from the domain.