Answer
$\frac{5}{13}$
Work Step by Step
Step 1. Letting $h^{-1}(\frac{12}{5})=tan^{-1}(\frac{12}{5})=t$, we have $tan(t)=\frac{12}{5}$ and $t$ in quadrant I.
Step 2. Form a right triangle with sides $x=5, y=12, r=\sqrt {x^2+y^2}=13$ with $|t|$ facing $y$.
Step 3. We have $g(h^{-1}(\frac{12}{5}))=cos(t)=\frac{5}{13}$