Check the uniform convergence of series of functions

(i) We can use $M-$test here. Denote $s_k(x) = \sum_{i=1}^k (\frac{x}{i})^i$, then $$ |s_k(x) -s_j(x) | \le | \sum_{i=j+1}^k (\frac{\pi}{i})^i | \le |\sum_{i=j+1}^k (\frac{1}{2})^i |$$

for all $i, j \ge M > 2 \pi $.

(ii) Denote $s_k(x) = \sum_{i=1}^k \frac{1}{(x+\pi)^2 i^2 } $. Suppose $s_k \stackrel{u}\rightarrow s$, when you don't know the limit, use the Cauchy Formulation: For all $\varepsilon >0$ exists $M$, such that for all $j > k \ge M$, $x \in ( -\pi, \pi)$.

$$|s_k(x) - s_j(x)| = \sum_{i=j+1}^k \frac{1}{(x+\pi)^2i^2} < \varepsilon $$

This clearly does not hold, pick $x = -\pi + \frac{1}{k}$ then, $$ 1\le \sum_{i=j+1}^k \frac{1}{(x+\pi)^2 k^2} \le |s_k(x) - s_j(x)| .$$

You calculated the radius of convergence incorrectly for the first series. Note that

$$\sum_{n=1}^\infty \left(\frac xn\right)^n = \sum_{n=1}^\infty c_nx^n$$

where $c_n=\frac{1}{n^n}$.

This means that $$\frac{c_n}{c_{n+1}} = \frac{(n+1)^{n+1}}{n^n} = \frac{(n+1)^n}{n^n}\cdot (n+1)$$ and the limit if this is not $e$.