Answer
$\theta=\frac{(b^{m+n+1}-a^{m+n+1})\sqrt 2m+1\sqrt 2n+1}{\sqrt (b^{2m+1}-a^{2m+1})(b^{2n+1}-a^{2n+1})}$
Work Step by Step
Let us consider that $\theta$ be an angle between the vectors $f$ and $g$, which can be written as: $\cos \theta=\dfrac{\langle f,g\rangle}{\|g\|\|g\|}$
Since, we have
$\cos \theta=\frac{\frac{b^{m+n+1}-a^{m+n+1}}{m+n+1}}{\sqrt \frac{b^{2m+1}-a^{2m+1}}{2m+1}\sqrt \frac{b^{2n+1}-a^{2n+1}}{2n+1}}=\frac{(b^{m+n+1}-a^{m+n+1})\sqrt 2m+12n+1}{\sqrt (b^{2m+1}-a^{2m+1})(b^{2n+1}-a^{2n+1})}$
So, the angle between $f$ and $g$ is equal to $\theta=\frac{(b^{m+n+1}-a^{m+n+1})\sqrt 2m+1\sqrt 2n+1}{\sqrt (b^{2m+1}-a^{2m+1})(b^{2n+1}-a^{2n+1})}$
