Answer
$-\sqrt {15}$
Work Step by Step
Step 1. Letting $g^{-1}(-\frac{1}{4})=cos^{-1}(-\frac{1}{4})=t$, we have $cos(t)=-\frac{1}{4}$ and $t$ in quadrant II.
Step 2. Form a right triangle of sides $x=1, r=4, y=\sqrt {r^2-x^2}=\sqrt {15}$ with $\pi-|t|$ facing $y$.
Step 3. We have $h(g^{-1}(-\frac{1}{4}))=tan(t)=-\sqrt {15}$