Answer
See table and explanations.
Work Step by Step
(a) $f(g(x))=|g(x)|=|\frac{1}{x-1}|$
(b) $f(g(x))=\frac{g(x)-1}{g(x)}=\frac{x}{x+1}$, thus $xg(x)=xg(x)+g(x)-x-1$ which gives $g(x)=x+1$
(c) $f(g(x))=\sqrt {g(x)}=|x|$, we have $g(x)=|x|^2=x^2$
(d) $f(g(x))=f(\sqrt x)=|x|=|(\sqrt x)^2|$, thus $f(x)=|x^2|=x^2$
