14 A school's mathematics extracurricular activity group simulates a desert tree planting plan on coordinate paper as follows: The $k$-th tree is planted at point $P_{k}\left(x_{k}, y_{k}\right)$, where $x_{1}=1, y_{1}=1$, and when $k \geqslant 2$,
$$
\left\{\begin{array}{l}
x_{k}=x_{k-1}+1-5\left[\frac{k-1}{5}\right]+5\left[\frac{k-2}{5}\right], \\
y_{k}=y_{k-1}+\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right] .
\end{array}\right.
$$

where $[a]$ represents the integer part of the real number $a$, for example $[2.6]=2,[0.6]=0$. According to this plan, the coordinates of the 2008th tree planting point are $\qquad$
(14) Let
$$
f(k)=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right],
$$

then
$$
\begin{aligned}
f(k+5) & =\left[\frac{k+5-1}{5}\right]-\left[\frac{k+5-2}{5}\right] \\
& =\left[1+\frac{k-1}{5}\right]-\left[1+\frac{k-2}{5}\right] \\
& =\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right] \\
& =f(k)
\end{aligned}
$$

Thus, $f(k)$ is a periodic function with a period of 5.
Calculations show: $f(2)=0, f(3)=0, f(4)=0, f(5)=0, f(6)=1$. Therefore,
$$
\begin{array}{l} 
x_{2008}=x_{2007}+1-5 f(2008), \\
x_{2007}=x_{2006}+1-5 f(2007), \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
x_{2}=x_{1}+1-5 f(2) .
\end{array}
$$

Adding up the above equations, we get
$$
\begin{aligned}
x_{2008} & =x_{1}+2007-5[f(2)+f(3)+\cdots+f(2008)] \\
& =x_{1}+2007-5\{401[f(2)+f(3)+\cdots+f(6)]+f(2)+f(3)\} \\
& =x_{1}+2007-5 \times 401 \\
& =3 .
\end{aligned}
$$

Similarly, we get $y_{2008}=402$. Therefore, the planting point of the 2008th tree is $(3,402)$.