2. Find all natural values of $n$ for which $\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9}$, and $\log _{3}^{2} n+14<\log _{3} 9 n^{7}$.
Solution: $\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9} \Leftrightarrow$ $2 \sin \frac{\pi}{9} \cos \frac{2 \pi}{9}+2 \sin \frac{\pi}{9} \cos \frac{4 \pi}{9}+\cdots+2 \sin \frac{\pi}{9} \cos \frac{2 \pi n}{9}=2 \sin \frac{\pi}{9} \cos \frac{\pi}{9} \Leftrightarrow$ $\sin \frac{3 \pi}{9}-\sin \frac{\pi}{9}+\sin \frac{5 \pi}{9}-\sin \frac{3 \pi}{9}+\cdots+\sin \frac{\pi}{9}(1+2 n)-\sin \frac{\pi}{9}(2 n-1)=\sin \frac{2 \pi}{9} \Leftrightarrow$ $\sin \frac{\pi}{9}(1+2 n)-\sin \frac{\pi}{9}=\sin \frac{2 \pi}{9} \Leftrightarrow \sin \frac{\pi}{9}(1+2 n)=\sin \frac{2 \pi}{9}+\sin \frac{\pi}{9} \Leftrightarrow \sin \frac{\pi}{9}(1+2 n)=2 \sin \frac{\pi}{6} \cos \frac{\pi}{18} \Leftrightarrow$ $\sin \frac{\pi}{9}(1+2 n)=\sin \frac{4 \pi}{9} \Leftrightarrow 2 \sin \left(\frac{\pi n}{9}-\frac{\pi}{6}\right) \cos \left(\frac{\pi n}{9}+\frac{5 \pi}{18}\right)=0 \Leftrightarrow$ $\frac{\pi n}{9}-\frac{\pi}{6}=\pi k, k \in Z, \quad \frac{\pi n}{9}+\frac{5 \pi}{18}=\frac{\pi}{2}+\pi m, m \in Z, \Leftrightarrow 2 n=3(6 k+1), k \in Z, \quad n=2+9 m, m \in Z$.

Since $n$ is a natural number, the first relation does not hold (an even number cannot be equal to an odd number). Therefore, $n=2+9 m, m \in Z, m \geq 0$.

Solve the inequality $\log _{3}^{2} n+14<\log _{3} 9 n^{7} \Leftrightarrow \log _{3}^{2} n-7 \log _{2} n+12<0, \Leftrightarrow\left(\log _{3} n-3\right)\left(\log _{3} n-4\right)<0, \Leftrightarrow$ $3^{3}<n<3^{4}, \Leftrightarrow 27<2+9 m<81, m \in Z, m \geq 0 \Leftrightarrow 2 \frac{7}{9}<m<8 \frac{7}{9}, m \in Z, m \geq 0 \Leftrightarrow m=3,4,5,6,7,8$.

Answer: $n=2+9 m, m=3,4,5,6,7,8$.