Solution

Let \[ \sin^{-1}\left(\frac{5}{13}\right) = x \]

Then, \[ \sin x = \frac{5}{13} \] \[ \cos x = \sqrt{1-\sin^2 x} = \frac{12}{13} \]

Therefore, \[ \tan x = \frac{\sin x}{\cos x} = \frac{5}{12} \]

Let \[ \cos^{-1}\left(\frac{3}{5}\right) = y \]

Then, \[ \cos y = \frac{3}{5} \] \[ \sin y = \sqrt{1-\cos^2 y} = \frac{4}{5} \]

Therefore, \[ \tan y = \frac{4}{3} \]

Using identity, \[ \tan(x+y) = \frac{\tan x + \tan y} {1-\tan x \tan y} \]

\[ = \frac{\frac{5}{12}+\frac{4}{3}} {1-\frac{5}{12}\cdot\frac{4}{3}} \]

\[ = \frac{\frac{63}{36}}{\frac{16}{36}} = \frac{63}{16} \]

Hence, \[ x+y = \tan^{-1}\left(\frac{63}{16}\right) \]

\[ \boxed{ \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) } \]