Let $\theta=\tan^{-1}\left(\frac{4}{3}\right).$ Since inverse trigonometric functions take principal values, we have $\theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right),$ and because $\frac{4}{3}>0$, therefore $\theta$ is in the first quadrant.

Now use the right triangle corresponding to $\tan\theta=\frac{4}{3}.$ So we can take:

• opposite side $=4$
• adjacent side $=3$
• hypotenuse $=5$

Hence, $\sin\theta=\frac{4}{5}, \qquad \cos\theta=\frac{3}{5}.$

Substitute into the given expression inside $\cos^{-1}$: $\frac{3}{10}\cos\left(\tan^{-1}\frac{4}{3}\right)+\frac{2}{5}\sin\left(\tan^{-1}\frac{4}{3}\right)$ $=\frac{3}{10}\cdot \frac{3}{5}+\frac{2}{5}\cdot \frac{4}{5}.$

Simplify: $=\frac{9}{50}+\frac{8}{25}$ $=\frac{9}{50}+\frac{16}{50}$ $=\frac{25}{50}=\frac{1}{2}.$

Therefore the whole expression becomes $\cos^{-1}\left(\frac{1}{2}\right).$

Since principal value range of $\cos^{-1}x$ is $[0,\pi]$, $\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}.$

So the correct option is $\boxed{\frac{\pi}{3}}.$