Correct Answer: $\frac{8 \sqrt{3}+1}{15}$

Solution : $\sin A = \frac{1}{3}$
⇒ $\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-(\frac{1}{9})} = \frac{\sqrt8}{3}$
Also, $\sin B = \frac{1}{5}$
⇒ $\cos B = \sqrt{1-\sin^2 B} = \sqrt{1-(\frac{1}{25})} = \frac{2\sqrt6}{5}$
Now, $\cos(A-B) = \cos A \cos B + \sin A \sin B$
$= \frac{\sqrt8}{3} × \frac{2\sqrt6}{5} + \frac{1}{3} × \frac{1}{5}$
$= \frac{8\sqrt3+1}{15}$
Hence, the correct answer is $\frac{8\sqrt3+1}{15}$.