Answer
$\dfrac{-3}{4}$
Work Step by Step
We know that
$ g(x)=\cos(x), h(x)=\tan(x)$
Thus we can write:
$h(g^{-1} (\dfrac{ -4}{5})=\tan (\cos^{-1}(\dfrac{ -4}{5})$
Suppose that $-\theta=\cos^{-1} (\dfrac{-4}{5})$
We know from SOH-CAH-TOA:
$\cos {\theta}=\dfrac{adjacent}{hypotenuse}=\dfrac{4}{5}$
To find the missing opposite side, we use the Pythagorean Theorem:
$opposite^2+4^2=5^2$
$opposite=\sqrt{(5)^2-(4)^2}=\sqrt {9}=3$
Therefore,
$\tan (-\theta)=-\dfrac{opposite}{adjacent}=\dfrac{-3}{4}$
Note that we are in quadrant II because cosine is negative, so the tangent of the angle should be negative.