8. Given the function $f:\{0,1, \cdots, 2010\} \rightarrow \mathbf{N}$. If for all possible integers $x$, we have
$$
\begin{array}{l}
f(4 x+2)=f(4 x+1), \\
f(5 x+3)=f(5 x+2), \\
f(7 x+5)=f(7 x+4),
\end{array}
$$

then $f(x)$ can take at most $\qquad$ different values.
8. 1033 .

For the function $f$, construct the function $g$, defined as:
$$
\begin{array}{l}
g(0)=1, \\
g(i+1)=\left\{\begin{array}{l}
1, f(i+1) \neq f(i) ; \\
0, f(i+1)=f(i),
\end{array}\right.
\end{array}
$$

where $i=0,1, \cdots, 2009$.
Then $g$ is a mapping from $\{0,1, \cdots, 2010\}$ to $\{0,1\}$, and for all possible $x$, we have
$$
g(4 x+2)=g(5 x+3)=g(7 x+5)=0 .
$$

Thus, $f(x)$ can take at most $\sum_{i=0}^{2010} g(i)$ different values.
Let $g(x)=h(x+2)$. Then $h$ is a mapping from $\{2,3, \cdots, 2012\}$ to $\{0,1\}$, and for all possible $x$, we have
$$
\begin{array}{l}
h(4 x)=h(5 x)=h(7 x)=0 . \\
\text { Since }\left(\sum_{i=2}^{2012} h(i)\right)_{\max } \\
=2011-\left[\frac{2012}{4}\right]-\left[\frac{2012}{5}\right]-\left[\frac{2012}{7}\right]+ \\
\quad\left[\frac{2012}{4 \times 5}\right]+\left[\frac{2012}{4 \times 7}\right]+\left[\frac{2012}{5 \times 7}\right]-\left[\frac{2012}{4 \times 5 \times 7}\right] \\
=1033,
\end{array}
$$
$\left(\right.$ at this time, $h(i)=\left\{\begin{array}{l}0,4 \mid i \text { or } 5 \mid i \text { or } 7 \mid i ; \\ 1, \text { otherwise. }\end{array}\right)$ and $\sum_{i=0}^{2010} g(i)=\sum_{i=2}^{2012} h(i)$, so $f(x)$ can take at most 1033 different values.