11. 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 $a_{16}=4, a_{17}=1$, find $a_{2018}$;
(2) Prove: From $\left\{a_{n}\right\}$, it is always possible to select infinitely many terms to form two different constant subsequences.
Solution (1) Since $a_{16}=4, a_{17}=1, a_{18}=3, a_{19}=2, a_{20}=1, a_{21}=1, a_{22}=0, a_{23}=1, a_{24}=1, a_{25}=0, \cdots$, starting from the 20th term, every three consecutive terms cyclically take the values $1,1,0$. Also, $2018=19+666 \times 3+1$, so $a_{2018}=1$.
5 points
Proof (2) First, we prove that the sequence $\left\{a_{n}\right\}$ must have a "0" term after a finite number of terms.
Assume that $\left\{a_{n}\right\}$ does not contain any "0" terms. Since $a_{n+2}=\left|a_{n+1}-a_{n}\right|$, for $n \geq 3$, we have $a_{n} \geq 1$.
If $a_{n+1}>a_{n}$, then $a_{n+2}=a_{n+1}-a_{n} \leq a_{n+1}-1(n \geq 3)$;
If $a_{n+1}a_{2 n+2}\right) \\
a_{2 n+2} & \left(a_{2 n+1}0$ leads to a contradiction. Therefore, $\left\{a_{n}\right\}$ must contain a "0" term.

If the first "0" 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 $0, M, M$, i.e.,
$$
a_{n+3 k}=0, \quad a_{n+3 k+1}=M, \quad a_{n+3 k+2}=M, k=0,1,2
$$

Thus, the sequence $\left\{a_{n}\right\}$ can definitely select infinitely many terms to form two different constant subsequences.
20 points