jmc

Setting the expressions for $g(x)$ equal to each other, we get $5x-4=f^{-1}(x)-3$, so $f^{-1}(x)=5x-1$. If we substitute $f(x)$ into this equation for $x$, we get $$f^{-1}(f(x))=5f(x)-1.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $x=5f(x)-1$. Solving for $f(x)$, we find $$f(x)=\frac{x+1}{5}.$$Thus, $a=\frac{1}{5}$ and $b=\frac{1}{5}$, so $5a+5b=\boxed{2}$.