jmc

We know that $f^{-1}(u)=v$ is the same as $u=f(v)$. Therefore $f^{-1}(g(x))=5x+3$ is the same as $$g(x)=f(5x+3).$$We can also use that $g(s)=t$ is equivalent to $s=g^{-1}(t)$ to say $$x=g^{-1}(f(5x+3)).$$This gives an expression containing $g^{-1}\circ f$.

Now we solve: $$g^{-1}(f(-7))=g^{-1}(f(5(-2)+3)).$$If $x=-2$ the equation $g^{-1}(f(5x+3))=x$ tells us $$g^{-1}(f(5(-2)+3))=\boxed{-2}.$$