3. Given the sequence $\left\{x_{n}\right\}$, and $x_{n+1}=\frac{x_{n}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}$, then $x_{1001}-x_{401}=$
3. 0 Detailed Explanation: From the structural characteristics, we can make the substitution $x_{n}=\tan a_{n}, \therefore x_{n+1}=\frac{\tan a_{n}+\tan 15^{\circ}}{1-\tan a_{n} \cdot \tan 15^{\circ}}=\tan \left(a_{n}+15^{\circ}\right)$, similarly $x_{n+2}=\tan \left[\left(a_{n}+15^{\circ}\right)+15^{\circ}\right]=\tan \left(a_{n}+2 \cdot 15^{\circ}\right), x_{n+12}=\tan \left(a_{n}+12 \cdot 15^{\circ}\right)=\tan \left(a_{n}+180^{\circ}\right)=\tan a_{n}=$ $x_{n}, \therefore\left\{x_{n}\right\}$ has a period of $12, \because 1001=50 \times 12+401$, hence $x_{1001}-x_{401}=0$.