Answer
$f$ and $g$ are inverses of each other.
The vaules $x\lt2$ are to be excluded.
Work Step by Step
$f(g(x))=f(\sqrt{ x}+2)=((\sqrt{ x}+2)-2)^2=(\sqrt{ x})^2=x$
$g(f(x))=g((x-2)^2)=\sqrt{(x-2)^2}+2=x-2+2=x$
Since for $f(x)$, $x\geq2$, then $x\lt2$ are values to be excluded.