1) Considering that we need to use the half-angle formula for that angle:
[tex]\sin (\frac{\theta}{2})=\pm\sqrt[]{\frac{1-\cos(\theta)}{2}}[/tex]2) But before that, we need to find the cosine of theta using the Fundamental Trigonometric Identity:
[tex]\begin{gathered} \sin ^2(\theta)+\cos ^2(\theta)=1 \\ \cos (\theta)=\sqrt[]{1-\sin ^2(\theta)} \\ \cos (\theta)=\sqrt[]{1-(\frac{12}{13})^2_{}} \\ \cos (\theta)=\sqrt[]{1-\frac{144}{169}} \\ \cos (\theta)=\pm\sqrt[]{\frac{25}{169}}=\pm\frac{5}{13} \end{gathered}[/tex]Note that since the angle is in Quadrant II, then we can state that
the cosine of theta has a negative value: -5/13
2.2) Let's plug into that and find out the value of sine (theta/2):
[tex]\begin{gathered} \sin (\frac{\theta}{2})=\pm\sqrt[]{\frac{1-\cos(\theta)}{2}} \\ \sin (\frac{\theta}{2})=\pm\sqrt[]{\frac{1-(-\frac{5}{13})}{2}}=\frac{3\sqrt[]{13}}{2} \end{gathered}[/tex]And that is the answer