let $\theta = \text{ arccsc}\left(\dfrac{13}{5}\right)$, assume $0<\theta<\dfrac{\pi}{2}$
$\sin(2\theta) = 2\sin{\theta} \cos{\theta}$
sketch a reference triangle in quad I (see attach)
$2 \cdot \dfrac{5}{13} \cdot \dfrac{12}{13} = \dfrac{120}{169}$
... have you studied identities yet?
As I said in another reply: you are not prepared to do these questions.
You need to learn(memorize) these sum/difference and double angle formula.
$\sin(2A)=2\sin(a)\cos(A)~\&~\cos(2A)=\left\{ \begin{array}{l}\cos^2(A)-\sin^2(A)\\1-2\sin^2(A)\\2\cos^2(A)-1\end{array} \right.$