11. (15 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}, a_{2}$ are given non-zero integers, $a_{n+2}=\left|a_{n+1}-a_{n}\right|$.
（1）If $a_{15}=2, a_{16}=-1$, find $a_{2008}$;
（2）Prove: From $\left\{a_{n}\right\}$, it is always possible to select infinitely many terms to form two different constant sequences.
$$
\begin{array}{l}
\text { 11. Solution: (1) } \because a_{15}=2, a_{16}=-1, a_{17}=3, a_{18}= \\
4, a_{19}=1, a_{20}=3, a_{21}=2, a_{22}=1, a_{23}=1, a_{24}=0, a_{25} \\
=1, a_{26}=1, a_{27}=0, \cdots
\end{array}
$$
11. Solution: (1) $\because a_{15}=2, a_{16}=-1, a_{17}=3, a_{18}=$ $4, a_{19}=1, a_{20}=3, a_{21}=2, a_{22}=1, a_{23}=1, a_{24}=0, a_{25}$ $=1, a_{26}=1, a_{27}=0, \cdots$
$\therefore$ Starting from the 22nd term, every three consecutive terms cyclically take the values $1,1,0$, hence $a_{2008}=1$.
(2) First, we prove that the sequence $\left\{a_{n}\right\}$ must have a zero term after a finite number of terms. Assume $\left\{a_{n}\right\}$ contains no zero terms,

Since $a_{n+2}=\left|a_{n+1}-a_{n}\right|$, for $n \geqslant 3$, we have $a_{n}$ $\geqslant 1$.

When $a_{n+1}>a_{n}$, $a_{n+2}=a_{n+1}-a_{n} \leqslant a_{n+1}-1(n \geqslant$ 3);

When $a_{n+1}a_{2 n+2}\right) \\
a_{2 n+2} & \left(a_{2 n+1}0$ contradiction, thus $\left\{a_{n}\right\}$ must contain a zero term.

If the first zero term is $a_{n}$, let $a_{n-1}=M(M \neq$ 0 ), then starting from the $n$-th term, every three consecutive terms cyclically take the values
$$
0, M, M \text {, i.e., }\left\{\begin{array}{l}
a_{n+3 k}=0 \\
a_{n+3 k+1}=M, k=0,1,2, \cdots \\
a_{n+3 k+2}=M
\end{array}\right.
$$

Therefore, the sequence $\left\{a_{n}\right\}$ can definitely select infinitely many terms to form two different constant sequences.
$$