5. Let
$$
\begin{array}{l}
f(x)=\left(x+\frac{1-\sqrt{3} \mathrm{i}}{2}\right)^{2010} \\
=\sum_{k=0}^{2010} a_{k} x^{k}+\mathrm{i} \sum_{k=0}^{2010} b_{k} x^{k},
\end{array}
$$

where, $a_{k}, b_{k} \in \mathbf{R}, k=0,1, \cdots, 2010$. Then
$$
\sum_{k=0}^{670}\left(a_{3 k}+b_{3 k}\right)=
$$
$\qquad$
5. $-2 \times 3^{1004}$.

Notice that
$$
\begin{array}{l}
f(x)=\left(x+\mathrm{e}^{-\frac{\pi}{3} \mathrm{i}}\right)^{2010}=\left(1+\mathrm{e}^{\frac{\pi}{3} \mathrm{i}} x\right)^{2010} \\
\sum_{k=0}^{2010} \mathrm{C}_{2010}^{k} x^{k} \cdot \mathrm{e}^{-\frac{\pi}{3} \mathrm{i}},
\end{array}
$$

So $\sum_{k=0}^{670} b_{3 k}=0$, and
$$
\begin{array}{l}
\sum_{k=0}^{670} a_{3 k}=\frac{1}{3} \sum_{j=0}^{2} f\left(e^{j \frac{2 \pi}{3}} \mathrm{i}\right) \\
=\frac{1}{3}\left[\left(1+\mathrm{e}^{\frac{\pi}{3} \mathrm{i}}\right)^{2000}+0+\left(1+\mathrm{e}^{-\frac{\pi}{3} \mathrm{i}}\right)^{2000}\right] \\
=3^{1004}\left[\left(\mathrm{e}^{\frac{\pi}{6} i}\right)^{2010}+\left(\mathrm{e}^{-\frac{\pi}{6} i}\right)^{2010}\right] \\
=-2 \times 3^{1004} \text {. } \\
\end{array}
$$