The value of $$\tan\left(2\tan^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right)\right)$$ is equal to:

Let $$\alpha = 2\tan^{-1}\!\left(\frac{3}{5}\right)$$ and $$\beta = \sin^{-1}\!\left(\frac{5}{13}\right)$$. We compute $$\tan(\alpha + \beta)$$.

Using the double-angle formula: $$\tan\alpha = \tan\!\left(2\tan^{-1}\frac{3}{5}\right) = \frac{2 \cdot \frac{3}{5}}{1 - \left(\frac{3}{5}\right)^2} = \frac{\frac{6}{5}}{\frac{16}{25}} = \frac{6}{5} \cdot \frac{25}{16} = \frac{15}{8}.$$

For $$\beta = \sin^{-1}\!\left(\frac{5}{13}\right)$$: $$\sin\beta = \frac{5}{13}$$, so $$\cos\beta = \frac{12}{13}$$ and $$\tan\beta = \frac{5}{12}$$.

Applying the addition formula: $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}}.$$

$$\text{Numerator: } \frac{15}{8} + \frac{5}{12} = \frac{45}{24} + \frac{10}{24} = \frac{55}{24}.$$ $$\text{Denominator: } 1 - \frac{75}{96} = \frac{21}{96} = \frac{7}{32}.$$

$$\tan(\alpha + \beta) = \frac{55}{24} \div \frac{7}{32} = \frac{55}{24} \cdot \frac{32}{7} = \frac{55 \cdot 4}{3 \cdot 7} = \frac{220}{21}.$$