Given $\tan \alpha + \frac{1}{\tan \alpha} = \frac{10}{3}$, for $\alpha \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, we solve for $\tan\alpha$. Thus $\tan\alpha = 3$ or $\tan\alpha = \frac{1}{3}$ (discarded as it is not in the given interval).

We have
\begin{align*}
\sin\left(2\alpha + \frac{\pi}{4}\right) + 2\cos\frac{\pi}{4}\cos^2\alpha &= \sin(2\alpha)\cos\frac{\pi}{4} + \cos(2\alpha)\sin\frac{\pi}{4} + \sqrt{2}\cos^2\alpha \\
&= \frac{\sqrt{2}}{2}\sin(2\alpha) + \sqrt{2}\cos(2\alpha) + \frac{\sqrt{2}}{2}(\cos^2\alpha + \sin^2\alpha) \\
&= \frac{\sqrt{2}}{2}\cdot\frac{2\sin\alpha\cos\alpha}{\sin^2\alpha + \cos^2\alpha} + \sqrt{2}\cdot\frac{\cos^2\alpha - \sin^2\alpha}{\sin^2\alpha + \cos^2\alpha} + \frac{\sqrt{2}}{2} \\
&= \frac{\sqrt{2}}{2}\cdot\frac{2\tan\alpha}{\tan^2\alpha + 1} + \sqrt{2}\cdot\frac{1 - \tan^2\alpha}{\tan^2\alpha + 1} + \frac{\sqrt{2}}{2} \\
&= \frac{\sqrt{2}}{2}\cdot\frac{2 \cdot 3}{3^2 + 1} + \sqrt{2}\cdot\frac{1 - 3^2}{3^2 + 1} + \frac{\sqrt{2}}{2} \\
&= \frac{\sqrt{2}}{2}\cdot\frac{6}{10} + \sqrt{2}\cdot\frac{1 - 9}{10} + \frac{\sqrt{2}}{2} \\
&= 0.
\end{align*}

The result is $\boxed{0}$.