Given \(\alpha = \cos^{-1}\left(\frac{8}{17}\right)\), we have \(\cos \alpha = \frac{8}{17}\).

Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we find \(\sin \alpha = \sqrt{1 - \left(\frac{8}{17}\right)^2}\).

Calculate \(\sin \alpha = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}\).

Now, \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{15/17}{8/17} = \frac{15}{8}\).

We need to find \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha}\).

\(\frac{1}{\sin \alpha} = \frac{17}{15}\) and \(\frac{1}{\tan \alpha} = \frac{8}{15}\).

Adding these, \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha} = \frac{17}{15} + \frac{8}{15} = \frac{25}{15} = \frac{5}{3}\).