If $$\cot x=\frac{5}{12}$$ for some $$x\in \left(\pi,\frac{3\pi}{2}\right)$$, then $$\sin 7x \left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right)+\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right)$$ is equal to

Simplify the expression:
$$\sin7x(\cos\frac{13x}{2}+\sin\frac{13x}{2})+\cos7x(\cos\frac{13x}{2}-\sin\frac{13x}{2})$$

Group terms:
$$=(\sin7x+\cos7x)\cos\frac{13x}{2}+(\sin7x-\cos7x)\sin\frac{13x}{2}$$

This matches the identity:
$$\sin(7x-\frac{13x}{2})+\cos(7x-\frac{13x}{2})$$

So it becomes:
$$\sin\frac{x}{2}+\cos\frac{x}{2}$$

Given:
$$\cot x=\frac{5}{12},\quad x\in(\pi,\frac{3\pi}{2})$$

$$\sin x=-\frac{12}{13},\quad\cos x=-\frac{5}{13}$$

s$$in\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\frac{3}{\sqrt{13}},\quad$$
$$\cos\frac{x}{2}=-\sqrt{\frac{1+\cos x}{2}}=-\frac{2}{\sqrt{13}}$$

$$\sin\frac{x}{2}+\cos\frac{x}{2}=\frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}=\frac{1}{\sqrt{13}}$$