Solution:

$Let \sin ^{-1}\left(\frac{3}{5}\right)=A$

$\therefore \sin A=\frac{3}{5}$

$\therefore \cos ^{2} A=1-\sin ^{2} A$

$\quad=1-\frac{9}{25}$

$\quad=\frac{16}{25}$

$\therefore \cos A=\frac{4}{5}$

$\begin{aligned} \sin ^{-1}\left(\frac{8}{17}\right)=B \\ & \therefore \sin B=\frac{8}{17} \\ & \therefore \cos ^{2} B \\ &=1-\sin ^{2} B \\ & = 1-\frac{64}{289} \\ & = \frac{225}{289} \\ \therefore \cos B=& \frac{15}{17} \therefore \cos (A-B)\\ &=\cos A \cos B+\sin A \sin B \\ &=\frac{4}{5} \times \frac{15}{17}+\frac{3}{5} \times \frac{8}{17} \\ \therefore \cos (A-B) &=\frac{84}{85} \\ \therefore A-B=& \cos ^{-1}\left(\frac{84}{85}\right) \\ \sin ^{-1}\left(\frac{3}{5}\right)-\sin ^{-1}\left(\frac{8}{17}\right)=\cos ^{-1}\left(\frac{84}{85}\right) \end{aligned}$