Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
[3]
Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\begin{array}{l}
g(n)=\frac{1}{2} f(n) . \\
\text { Hence } a_{n-1}=1, a_{n}=2,3,4 ; \\
a_{n-1}=2, a_{n}=3,4 ; \\
a_{n-1}=3, a_{n}=4 .
\end{array}
$$

Let $m(i)$ represent the number of $(n-1)$-digit wave numbers where $a_{n-2}>a_{n-1}$ and the last digit is $i$. Then
$$
\begin{array}{l}
m(1)+m(2)+m(3)+m(4)=g(n-1), \\
m(4)=0 .
\end{array}
$$

When $a_{n-1}=1$, $a_{n-2}=2,3,4$. Then
$$
m(1)=g(n-2) \text {. }
$$

When $a_{n-1}=3$, $a_{n-2}=4, a_{n-3}=1,2,3$. Then $m(3)=g(n-3)$.
Thus, $g(n)=3 m(1)+2 m(2)+m(3)$
$$
=2 g(n-1)+g(n-2)-g(n-3) \text {. }
$$

It is easy to see that $g(2)=6, g(3)=14, g(4)=31$, $\cdots \cdots g(10)=4004$.
Therefore, $f(10)=2 g(10)=8008$.
Also note that the sequence of remainders when $g(n)$ is divided by 13 is:
$$
6,1,5,5,1,2,0,1,0,1,1,3,6,1,5,5, \cdots
$$

The smallest positive period is 12.
Thus, $g(2008) \equiv g(4) \equiv 5(\bmod 13)$.
Therefore, $f(2008) \equiv 10(\bmod 13)$.