5. $9 \sqrt{3}$.

From the problem, we know
$$
\begin{array}{l}
\boldsymbol{a}_{k+6} \\
=\left(\cos \left(\frac{k \pi}{6}+\pi\right), \sin \left(\frac{k \pi}{6}+\pi\right)+\cos \left(\frac{k \pi}{6}+\pi\right)\right) \\
=\left(-\cos \frac{k \pi}{6},-\sin \frac{k \pi}{6}-\cos \frac{k \pi}{6}\right) \\
=-\boldsymbol{a}_{k} \text {. } \\
\text { Therefore, } \sum_{k=0}^{11} \boldsymbol{a}_{k} \cdot \boldsymbol{a}_{k+1} \\
=a_{0} \cdot a_{1}+a_{1} \cdot a_{2}+a_{2} \cdot a_{3}+a_{3} \cdot a_{4}+a_{4} \cdot a_{5}+ \\
a_{5} \cdot a_{6}+a_{6} \cdot a_{7}+a_{7} \cdot a_{8}+a_{8} \cdot a_{9}+a_{9} \cdot a_{10}+ \\
a_{10} \cdot a_{11}+a_{11} \cdot a_{12} \\
=a_{0} \cdot a_{1}+a_{1} \cdot a_{2}+a_{2} \cdot a_{3}+a_{3} \cdot a_{4}+a_{4} \cdot a_{5}+ \\
a_{5} \cdot a_{6}+a_{0} \cdot a_{1}+a_{1} \cdot a_{2}+a_{2} \cdot a_{3}+a_{3} \cdot a_{4}+ \\
a_{4} \cdot a_{5}+a_{5} \cdot a_{6} \\
=2\left(a_{0} \cdot a_{1}+a_{1} \cdot a_{2}+a_{2} \cdot a_{3}+a_{3} \cdot a_{4}+a_{4} \cdot a_{5}+a_{5} \cdot a_{6}\right) \\
=9 \sqrt{3} \text {. } \\
\end{array}
$$
5. Let the vector
$$
\boldsymbol{a}_{k}=\left(\cos \frac{k \pi}{6}, \sin \frac{k \pi}{6}+\cos \frac{k \pi}{6}\right) \text {, }
$$

where, $k=0,1, \cdots, 12$. Then the value of $\sum_{k=0}^{11} \boldsymbol{a}_{k} \cdot \boldsymbol{a}_{k+1}$ is