jmc

We have $g(4)=\sqrt{2(4)+1}=\sqrt{9}=3$, so $f(g(4))=f(3)=3^2-2(3)+1=4$. Since $f(3)=4$, we have $g(f(3))=g(4)=\sqrt{2(4)+1}=3$. Therefore, $f(g(4))-g(f(3))=4-3=\boxed{1}$.

Notice that $f(g(4))=4$ and $g(f(3))=3$. Is that a coincidence?