(7) Given the sequence $\left\{a_{n}\right\}$ where the first term $a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}$, then $a_{2011}=$
(7) 2 Hint: $a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{13-7 a_{n}}{5-3 a_{n}}$,
$$
a_{n+3}=\frac{13-7 a_{n+1}}{5-3 a_{n+1}}=\frac{13-7 \cdot \frac{5 a_{n}-13}{3 a_{n}-7}}{5-3 \cdot \frac{5 a_{n}-13}{3 a_{n}-7}}=a_{n} \text {. }
$$

Therefore,
$$
a_{2011}=a_{3 \times 670+1}=a_{1}=2 .
$$