8. A school's mathematics extracurricular activity group designed a tree planting plan on a coordinate paper for a certain desert 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 $k \geqslant 2$, 
\[
\left\{
\begin{array}{l}
x_{k}=x_{k-1}+1-5\left[\frac{k-1}{5}\right]+5\left[\frac{k-2}{5}\right] ; \\
y_{k}=y_{k-1}+\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right] .
\end{array}
\right.
\]
Here, $[a]$ 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 $\qquad$
8. $(3,402)$ Detailed Explanation: Let $f(k)=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]$, then $f(k+5)=\left[\frac{k+5-1}{5}\right]-\left[\frac{k+5-2}{5}\right]=$ $\left[1+\frac{k-1}{5}\right]-\left[1+\frac{k-2}{5}\right]=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]=f(k)$, hence $f(k)$ is a function with a period of 5. Calculation shows: $f(2)=0$; $f(3)=0 ; f(4)=0 ; f(5)=0 ; f(6)=1$. Therefore, $x_{2008}=x_{2007}+1-5 f(2008) ; x_{2007}=x_{2006}+1-5 f(2007) ; \cdots$; $x_{2}=x_{1}+1-5 f(2)$. Adding all these equations, we get $x_{2008}=x_{1}+2007-5[f(2)+f(3)+\cdots+f(2008)]=x_{1}+2007-$ $5\{401[f(2)+f(3)+\cdots+f(6)]+f(2)+f(3)\}=x_{1}+2007-5 \times 401=3$; Similarly, we can get $y_{2008}=402$. Therefore, the planting point of the 2008th tree is $(3,402)$.