Since $\cos A=\large\frac{1}{7}$ & $\cos B=\large\frac{13}{14}$
$\therefore \sin A=\sqrt{1-\cos A}$
$\Rightarrow \sqrt{1-(\large\frac{1}{7})^2}=\large\frac{4\sqrt 3}{7}$
$\sin B=\sqrt{1-\cos^2B}=\sqrt{1-(\large\frac{13}{15})^2}=\large\frac{3\sqrt 3}{14}$
$\therefore\cos(A-B)=\cos A\cos B+\sin A.\sin B$
$\Rightarrow \large\frac{1}{7}.\frac{13}{14}+\frac{4\sqrt 3}{7}.\frac{3\sqrt 3}{14}$
$\Rightarrow \large\frac{49}{7\times 14}=\frac{1}{2}$
Hence (A) is the correct answer.