$$\begin{aligned} &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{r=1}^{13} \frac{\sin \left[\left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)-\left(\frac{\pi}{4}\right)-(\mathrm{r}-1) \frac{\pi}{6}\right]}{\sin \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)} \\ & \frac{1}{\sin \frac{\pi}{6}} \sum_{\mathrm{r}=1}^{13}\left(\cot \left(\frac{\pi}{4}+(\mathrm{r}-1) \frac{\pi}{6}\right)-\cot \left(\frac{\pi}{4}+\frac{\mathrm{r} \pi}{6}\right)\right) \\ & =2 \sqrt{3}-2=\alpha \sqrt{3}+\mathrm{b} \end{aligned}\\ &\text { So } \mathrm{a}^2+\mathrm{b}^2=8 \end{aligned}$$