Answer
$\dfrac{\pi}{6}$
Work Step by Step
We know that
$ f(x)=\sin(x), g(x)=\cos(x)$
Thus we can write:
$g^{-1}(f (\dfrac{-4\pi}{3}))=\cos^{-1} (\sin (\dfrac{-4 \pi}{3}))$
From the unit circle, we know that:
$\sin (\dfrac{-4\pi}{3})=\dfrac{-1}{2}$
Therefore,
$\cos^{-1} (\sin (\dfrac{-4 \pi}{3}))=\cos^{-1}(\dfrac{-1}{2})=\dfrac{\pi}{6}$
Note that we restrict the range of $\cos^{-1}$ to $[0,\pi]$.