1. The problem asks for the average power of the signal $x[n] = \cos\left(\frac{\pi n}{3}\right)$. 2. The average power $P$ of a discrete-time signal $x[n]$ is defined as $$P = \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N |x[n]|^2.$$ 3. For a cosine signal $x[n] = A \cos(\omega n + \phi)$, the average power is known to be $\frac{A^2}{2}$ because the square of cosine averages to $\frac{1}{2}$ over one or more periods. 4. In this problem, $A=1$, $\omega = \frac{\pi}{3}$, and $\phi=0$. 5. Therefore, the average power is $$P = \frac{1^2}{2} = \frac{1}{2}.$$ 6. This means the average power of $\cos\left(\frac{\pi n}{3}\right)$ is $0.5$.