Answer
See below.
Work Step by Step
$f\circ g=f(g(x))=f(\frac{x+5}{x-3})=\dfrac{\frac{x+5}{x-3}}{\frac{x+5}{x-3}-2}=\dfrac{\frac{x+5}{x-3}}{\frac{x+5}{x-3}-\frac{2(x-3)}{x-3}}=\frac{x+5}{x+5-2(x-3)}=\frac{x+5}{x+5-2x+6}=\frac{x+5}{11-x}$
The denominator cannot be 0, hence if $11-x=0$ and $x-3=0$ then $x=11,3$. Hence the domain is all real numbers except $11$ and $3$.
