But $\sin (360 - \theta) \ne \sin\theta$.
so you can not figure that if $\sin \theta=\frac 5{13};0\le \theta < 360^{\circ}$, that that would mean the reflexive angle $\psi = 360-\theta$ will have $\sin \psi = \sin \theta$. In fact just the opposite; $\sin \psi = -\sin \theta$.
So:
$\sin \theta = \frac 5{13}$.
Find $\theta$: $\frac 5{13}>$ so $0< \theta < 180$. So if a) $\theta$ is accute then $\theta = \arcsin \frac 5{13}$. If b) $\theta$ is reflexive this is not possible and there is no solution.
$\cos \theta = \pm \sqrt{1-\sin^2 \theta} = \pm\frac {12}{13}$. So if a) $\theta$ is accute then $\cos \theta \ge 0$ and $\cos \theta = \frac {12}{13}$. If b) then $\theta$ is impossible so there is no solution.
$\tan \theta = \frac {\sin \theta}{\cos \theta}$. So if a) $\theta$ is accute then $\tan \theta = \frac {\frac 5{13}}{\frac {12}{13}} = \frac 5{13}$ and, again, for (b) there is no solution.
Now it's not sure if I'm supposed to leave it as $\theta = \arcsin \frac {5}{13}$ or if I'm supposed to do some fancy geometry and trigonometry to find out what the angle of right triangle with sides of $5,12,13$ is. But I'm not going to grub in the mud unless I know there's a potato down there.
I could punch keys into a calculator to find $\theta = \arcsin \frac {5}{13}$ but nobody wants some stupid decimal when $\arcsin \frac {5}{13}$ is much more descriptive than $\theta = 22.6whogivesapileoffetiddingokidneys1087668.....$
So $\theta = \arcsin \frac 5{13}=\arctan \frac 5{12}$. Final answer.
.....
I suspect the person asking the question meant obtuse.
In which case b) $90 < \theta < 180$ and as $\sin{180 - \theta}=\sin \theta$ we have $\theta = 180 -\arcsin \theta$. And $\cos \theta < 0$ so $\cos \theta =-\frac 5{12}$ and $\tan \theta =-\frac 5{12}$. By the way we can express $\theta = \arctan -\frac 5{12}$.