8. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{n}=a_{n-1}-a_{n-2}, n \geqslant 3$.

If the sum of its first 1492 terms is 1985, and the sum of the first 1985 terms is 1492, find the sum of its first 2001 terms.
8. For any natural number $n$, by the assumption we have
$$
\begin{array}{l}
a_{n}, a_{n+1}, a_{n+2}=a_{n+1}-a_{n}, a_{n+3}=a_{n+2}-a_{n+1}=-a_{n}, \\
a_{n+4}=a_{n+3}-a_{n+2}=-a_{n+1}, a_{n+5}=a_{n}-a_{n+1}, a_{n+6}=a_{n} .
\end{array}
$$

From this, we can also get $a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0$.
Let the sum of the first $n$ terms of the sequence be $S_{n}$. Since $1492=6 \times 248+4,1985=6 \times 330+5$, we have
$$
\begin{array}{l}
1985=S_{1492}=a_{1}+a_{2}+a_{3}+a_{4}=2 a_{2}-a_{1}, \\
1492=S_{1985}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=a_{2}-a_{1} .
\end{array}
$$

Solving these, we get $a_{1}=-999, a_{2}=493$. Since $2001=6 \times 333+3$, we have $S_{2001}=a_{1}+a_{2}+a_{3}=2 a_{2}=986$.