Recall that
[tex]\cos^2x+\sin^2x=1[/tex]
which means
[tex]\sin^2x=1-\left(\dfrac5{13}\right)^2[/tex]
[tex]\sin x=\pm\sqrt{\dfrac{144}{169}}=\pm\dfrac{12}{13}[/tex]
and since [tex]\sin x<0[/tex] you take the negative root:
[tex]\sin x=-\dfrac{12}{13}[/tex]
Now, you know that [tex]\sin x<0[/tex] for [tex]\pi<x<2\pi[/tex] (third and fourth quadrants). In terms of [tex]\dfrac x2[/tex], you have [tex]\dfrac\pi2<\dfrac x2<\pi[/tex] (second quadrant). Over this interval, you have [tex]\sin\dfrac x2>0[/tex].
With this in mind, recall the half-angle identity for sine:
[tex]\sin^2x=\dfrac{1-\cos2x}2[/tex]
[tex]\sin x=\pm\sqrt{\dfrac{1-\cos2x}2}[/tex]
Replacing [tex]x[/tex] with [tex]\dfrac x2[/tex] gives
[tex]\sin\dfrac x2=\pm\sqrt{\dfrac{1-\cos x}2}[/tex]
and using the fact that [tex]\sin\dfrac x2>0[/tex], you take the positive root.
Finally, you have
[tex]\sin\dfrac x2=\sqrt{\dfrac{1-\frac5{13}}2}[/tex]
[tex]\sin\dfrac x2=\dfrac2{\sqrt{13}}[/tex]
