Let $\theta = \sin^{-1}\left(\frac{5}{13}\right)$ and $\phi = \cos^{-1}\left(\frac{3}{5}\right)$. Then, $\sin \theta = \frac{5}{13}$, and using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:

$$\cos \theta = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$

Hence,

$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$$

Similarly, $\cos \phi = \frac{3}{5}$, so:

$$\sin \phi = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

Thus,

$$\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$$

Now, using the tangent addition formula:

$$\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$$

Substituting the values:

$$\tan(\theta + \phi) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \left(\frac{5}{12}\right)\left(\frac{4}{3}\right)} = \frac{\frac{5}{12} + \frac{16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{1 - \frac{5}{9}} = \frac{\frac{7}{4}}{\frac{4}{9}} = \frac{7}{4} \times \frac{9}{4} = \frac{63}{16}$$

Since $\theta$ and $\phi$ are both in the first quadrant ($0 < \theta, \phi < \frac{\pi}{2}$), their sum $\theta + \phi$ lies between $0$ and $\pi$. Moreover, $\theta \approx 22.6°$ and $\phi \approx 53.1°$, so $\theta + \phi \approx 75.7°$, which is less than $\frac{\pi}{2}$ ($90°$). Therefore, $\theta + \phi$ is within the principal range of $\tan^{-1}$, i.e., $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Consequently:

$$\theta + \phi = \tan^{-1}\left(\frac{63}{16}\right)$$

Hence, $\sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right)$, as required.