12. Given a function $f(x)$ defined on $\mathbf{R}$ whose graph is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, and satisfies $f(x)=-f\left(x+\frac{3}{2}\right), f(-1)=1, f(0)=-2$, then the value of $f(1)+f(2)+\cdots+f(2008)$ is $\qquad$
12. From the fact that the graph of the function $f(x)$ is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, we know that $f(x)=-f\left(-x-\frac{3}{2}\right)$.

Given that $f(x)=-f\left(x+\frac{3}{2}\right)$, it follows that $f\left(-x-\frac{3}{2}\right)=f\left(x+\frac{3}{2}\right)$.
Therefore, $f(-x)=f(x)$.
Hence, $f(x+3)=f\left[\left(x+\frac{3}{2}\right)+\frac{3}{2}\right]=-f\left(x+\frac{3}{2}\right)=f(x)$.
Thus, $f(x)$ is an even function with a period of 3.
Consequently, $f(1)=f(-1)=1, f(2)=f(-1+3)=f(-1)=1, f(3)=f(0)=-2$.
Therefore, $f(1)+f(2)+\cdots+f(2008)=669[f(1)+f(2)+f(3)]+f(2008)=f(2008)=f(1)=1$.