Answer
(a) $f^{-1}(x) = \frac{2}{x-5}$
(b) Both $f(f^{-1}(x))$ and $f^{-1}(f(x))$ result in $x$.
Work Step by Step
(a) $$f(x) = \frac{2}{x} + 5$$ $$y = \frac{2}{x} + 5$$ $$x = \frac{2}{y} + 5$$ $$(x - 5) = \frac{2}{y}$$ $$y = \frac{2}{(x - 5)}$$ $$f^{-1}(x) = \frac{2}{x-5}$$
(b)
$f(f^{-1}(x))$ $$\frac{2}{\frac{2}{x-5}} + 5$$ $$x - 5 + 5$$ $$x$$
$f^{-1}(f(x))$ $$\frac{2}{(\frac{2}{x} + 5) - 5}$$ $$\frac{2}{\frac{2}{x}}$$ $$x$$
