Let $\theta = \tan^{-1}\left(\frac{3}{5}\right), \qquad \phi = \sin^{-1}\left(\frac{5}{13}\right).$

We need to find $\tan(2\theta + \phi).$

First compute $\tan 2\theta$.

Since $\tan \theta = \frac{3}{5},$ we use $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}.$

Therefore,

Now compute $\tan \phi$.

Given $\sin \phi = \frac{5}{13}.$ Since $\phi = \sin^{-1}(5/13)$ lies in the principal range $[-\pi/2,\pi/2]$, and $5/13>0$, we have $\phi$ in the first quadrant.

Hence $\cos \phi = \sqrt{1-\sin^2\phi} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}.$

So, $\tan \phi = \frac{\sin\phi}{\cos\phi} = \frac{5/13}{12/13} = \frac{5}{12}.$

Use the tangent addition formula: $\tan(2\theta+\phi) = \frac{\tan 2\theta + \tan \phi}{1-\tan 2\theta\tan \phi}.$

Substitute the values: $\tan(2\theta+\phi) = \frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\cdot \frac{5}{12}}.$

Simplify the numerator: $\frac{15}{8}+\frac{5}{12} = \frac{45+10}{24} = \frac{55}{24}.$

Simplify the denominator: $1-\frac{15}{8}\cdot\frac{5}{12} = 1-\frac{75}{96} = 1-\frac{25}{32} = \frac{7}{32}.$

Therefore, $\tan(2\theta+\phi) = \frac{55/24}{7/32} = \frac{55}{24}\cdot \frac{32}{7} = \frac{55\cdot 4}{3\cdot 7} = \frac{220}{21}.$

Hence the correct option is $\boxed{\frac{220}{21}}$ which is Option B.