Solution:

$ x(n)=e^{j\left(\frac{\pi}{2} n+\frac{\pi}{4}\right)}\\ $

To find Energy of x(n):

$ \begin{aligned} E & =\sum_{n=-\infty}^{\infty}|x(n)|^2 \\\\ & =\sum_{n=-\infty}^{\infty} \mid e^{\left.j\left(\frac{\pi}{2} n+\frac{\pi}{4}\right)\right|^2} u(n)\\ \\ & =\sum_{n=-\infty}^{\infty}\left|e^{\left.j\left(\frac{\pi}{2} n+\frac{\pi}{4}\right)\right|^2} \quad \because\right| \\e^{j(\omega+\theta)} \mid=1 \\\\ & =\sum_{n=-\infty}^{\infty} 1=\infty \end{aligned} $

To find power of x(n):

$ \begin{aligned} P & =\lim _{n \rightarrow \infty} \frac{1}{2 N+1} \sum_{n=-N}^N|x(n)|^2 \\\\ & =\lim \frac{1}{2 N+1} \sum_{n=-N}^N\left|e^{\left.j\left(\frac{\pi}{2} n+\frac{\pi}{4}\right)\right|^2}\right|^2 \\\\ & =\lim \frac{1}{2 N+1} \sum_{n=-N}^N 1 \\\\ & =\lim \frac{1}{2 N+1}(2 N+1) \\\\ & =1\\ \end{aligned}\\ $