(1) It is easy to see,
$$\begin{array}{l} a_{16}=4, a_{17}=1, a_{18}=3, a_{19}=2, a_{20}=1, \\ a_{21}=1, a_{22}=0, a_{23}=1, a_{24}=1, a_{25}=0, \\ \cdots \cdots \end{array}$$

Notice that, starting from the 20th term, every three consecutive terms periodically take the values $1, 1, 0$.
$$\begin{array}{l} \text { Also, } 2018=19+666 \times 3+1 \text {, so } \\ a_{2018}=1 . \end{array}$$
(2) First, we prove: the sequence $\left\{a_{n}\right\}$ must have a "0" term after a finite number of terms.
Assume $\left\{a_{n}\right\}$ does not have a "0" term.
Since $a_{n+2}=\left|a_{n+1}-a_{n}\right|$, then for $n \geqslant 3$, we have $a_{n} \geqslant 1$.
If $a_{n+1}>a_{n}$, then
$$a_{n+2}=a_{n+1}-a_{n} \leqslant a_{n+1}-1(n \geqslant 3) \text {; }$$

If $a_{n+1}a_{2 n+2} \\
a_{2 n+2}, & a_{2 n+1}0$ leads to a contradiction.
Therefore, $\left\{a_{n}\right\}$ must have 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 periodically take the values $0, M, M$, i.e.,
$$a_{n+3 k}=0, a_{n+3 k+1}=M, a_{n+3 k+2}=M,$$

where $k=0,1, \cdots$.
Thus, the sequence $\left\{a_{n}\right\}$ can select infinitely many terms to form two different constant subsequences.