Answer
See explanations.
$x\ne2$, $x\ne-4$
Work Step by Step
Step 1. Given $f(x)=\frac{2x+3}{x+4}$ and $g(x)=\frac{4x-3}{2-x}$, we have $f(g(x))=\frac{2(\frac{4x-3}{2-x})+3}{(\frac{4x-3}{2-x})+4}=\frac{8x-6+6-3x}{4x-3+8-4x}=x$ with $x\ne2$
Step 2. We have $g(f(x))=\frac{4(\frac{2x+3}{x+4})-3}{2-(\frac{2x+3}{x+4})}=\frac{8x+12-3x-12}{2x+8-2x-3}=x$ with $x\ne-4$
Step 3. Thus $f$ and $g$ are inverse functions.
Step 4. For $f(x)$, we have $x\ne-4$ and for $g(x)$, we have $x\ne2$ which are also part of the composite requirements.
