9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$

Find the value of $f(2012)$.
In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$.