10. Given the vectors $\boldsymbol{a}=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right), \boldsymbol{b}=(\cos \alpha, \sin \alpha), 0<\alpha<\frac{2 \pi}{3}$.
(1) Prove that $\boldsymbol{a}+\boldsymbol{b}$ is perpendicular to $\boldsymbol{a}-\boldsymbol{b}$;
(2) Let $\boldsymbol{x}=k \boldsymbol{a}+3 \boldsymbol{b}, \boldsymbol{y}=\boldsymbol{a}+\frac{1}{k} \boldsymbol{b}\left(k \in \mathbf{R}^{+}\right)$, if $\boldsymbol{x} \perp \boldsymbol{y}$, find the relationship between $k$ and $\alpha$;
(3) Under the condition of (2), determine the range of $\alpha$.
10. (1) Since $a=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right), b=(\cos \alpha, \sin \alpha)$, so $|\boldsymbol{a}|=1,|\boldsymbol{b}|=1$. Therefore, $(\boldsymbol{a}+ \boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=|\boldsymbol{a}|^{2}-|\boldsymbol{b}|^{2}=1-1=0$. Hence, $\boldsymbol{a}+\boldsymbol{b}$ is perpendicular to $\boldsymbol{a}-\boldsymbol{b}$.
(2) Since $\boldsymbol{x} \perp \boldsymbol{y}$, so $\boldsymbol{x} \cdot \boldsymbol{y}=0$. Also, $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{2} \cos \alpha-\frac{\sqrt{3}}{2} \sin \alpha=\sin \left(\frac{\pi}{6}-\alpha\right)$.
Therefore, $\boldsymbol{x} \cdot \boldsymbol{y}=(k \boldsymbol{a}+3 \boldsymbol{b}) \cdot\left(\boldsymbol{a}+\frac{1}{k} \boldsymbol{b}\right)=k|\boldsymbol{a}|^{2}+\frac{3}{k}|\boldsymbol{b}|^{2}+4 \boldsymbol{a} \cdot \boldsymbol{b}=k+\frac{3}{k}+4 \sin \left(\frac{\pi}{6}-\alpha\right)$. Hence, $k+\frac{3}{k}+4 \sin \left(\frac{\pi}{6}-\alpha\right)=0$.
(3) From (2), we know that $k+\frac{3}{k}+4 \sin \left(\frac{\pi}{6}-\alpha\right)=0$, so $k+\frac{3}{k}=4 \sin \left(\alpha-\frac{\pi}{6}\right)$. Since $k>0$, we have $k+\frac{3}{k} \geqslant 2 \sqrt{3}$, so $4 \sin \left(\alpha-\frac{\pi}{6}\right) \geqslant 2 \sqrt{3}$, which means $\sin \left(\alpha-\frac{\pi}{6}\right) \geqslant \frac{\sqrt{3}}{2}$. Given that $0<\alpha<\frac{2 \pi}{3}$, hence $\frac{\pi}{2} \leqslant \alpha<\frac{2 \pi}{3}$.