5. Product $\prod_{k=1}^{7}\left(1+2 \cos \frac{2 k \pi}{7}\right)=$ $\qquad$
5.3 .
$$
\begin{array}{l}
\text { Let } \omega=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7} \text {. Then } \omega^{7}=1 . \\
\omega^{k}=\cos \frac{2 k \pi}{7}+i \sin \frac{2 k \pi}{7}, \\
\omega^{-k}=\cos \frac{2 k \pi}{7}-i \sin \frac{2 k \pi}{7}, \\
\therefore \omega^{k}+\omega^{-k}=2 \cos \frac{2 k \pi}{7} . \\
\text { The original expression }=\prod_{k=1}^{7}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{7} \omega^{-k}\left(1+\omega^{k}+\omega^{2 k}\right) \\
=\omega^{-1 \cdots 2 \cdots \cdots-7} \cdot \prod_{k=1}^{6}\left(1+\omega^{k}+\omega^{2 k}\right) \cdot\left(1+\omega^{7}\right. \\
\left.\quad+\omega^{14}\right) \\
=\left(\omega^{7}\right)^{-4} \cdot 3 \prod_{k=1}^{6} \frac{1-\omega^{3 k}}{1-\omega^{k}} \\
=3 \cdot \frac{1-\omega^{3}}{1-\omega} \cdot \frac{1-\omega^{6}}{1-\omega^{2}} \cdot \frac{1-\omega^{9}}{1-\omega^{3}} \cdot \frac{1-\omega^{12}}{1-\omega^{4}} \cdot \frac{1-\omega^{15}}{1-\omega^{5}} \\
\cdot \frac{1-\omega^{18}}{1-\omega^{6}}=3 .
\end{array}
$$