(1) Let $\left(\frac{\sqrt{3}}{2}+\frac{x}{2} \mathrm{i}\right)^{2008}=f(x)+\mathrm{i} g(x)$, where $f(x) 、 g(x)$ are both real-coefficient polynomials, then the sum of the coefficients of $f(x)$ is ( ).
(A) $-\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{3}}{2}$
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}$
(1) Let $x=1$, then
$$
\left(\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}\right)^{2008}=f(1)+\mathrm{i} g(1),
$$

which means
$$
\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right)^{2008}=f(1)+\mathrm{i} g(1),
$$

thus
$$
\left(\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right)^{4}=f(1)+\mathrm{i} g(1),
$$

which is
$$
\begin{aligned}
\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3} & =f(1)+\mathrm{i} g(1) \\
& =-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i} .
\end{aligned}
$$

Therefore, the answer is $\mathrm{C}$.