Answer
(a) $\frac{2x+7}{2x+3}$, $(-\infty,-\frac{3}{2})U(-\frac{3}{2},\infty)$
(b) $ 5$
(c) $ -3$
Work Step by Step
(a) Given $f(x)=\frac{x+2}{x-2}$ and $g(x)=2x+5$, we have $f\circ g=f(g(x))=\frac{2x+5+2}{2x+5-2}=\frac{2x+7}{2x+3}$ with domain $x\ne -3/2$ or $(-\infty,-\frac{3}{2})U(-\frac{3}{2},\infty)$
(b) $(g\circ f)(-2)=g(f(-2))=2(\frac{-2+2}{-2-2})+5=5$
(c) $(f\circ g)=\frac{2(-2)+7}{2(-2)+3}=-3$
