$\cos \alpha=\frac{1}{13}$

We should first know the identity $\sin (\alpha + \beta) \sin (\alpha - \beta)=\sin^2 \alpha - \sin^2 \beta$ .

Now, given that $\cos \alpha = \frac{1}{13}$ and $\cos \beta = \frac{12}{13}$ . Then ----

$\sin (\alpha + \beta) \sin (\alpha - \beta)$

$=\sin^2 \alpha - \sin^2 \beta$

$=(1 - \cos^2 \alpha)-(1 - \cos^2 \beta)$ [since, $\sin^2 \theta = 1 - \cos^2 \theta$ ]

$=1 - cos^2 \alpha - 1 + \cos^2 \beta$

$=cos^2 \beta - \cos^2 \alpha$

$=(\cos \beta + \cos \alpha)(\cos \beta - \cos \alpha)$ [since, $a^2-b^2=(a+b)(a-b)$ ]

$=(\frac{12}{13}+\frac{1}{13})(\frac{12}{13}-\frac{1}{13})$

$=\frac{13}{13}\times \frac{11}{13} = 1\times \frac{11}{13} = \boxed{\frac{11}{13}}$

2.sinA.sinB=cos(A-B)-cos(A+B)

cos2A=2cos^2(A)-1

11/13