The first can be done without recourse to $\limsup$:
If $R$ is the radius of convergence of $\sum_{n=0}^{\infty}c_n x^n$, then the series converges absolutely if $|x|<R$ and diverges if $|x| > R$.
Hence $\sum_{n=0}^{\infty}c_n x^{2n}= \sum_{n=0}^{\infty}c_n (x^2)^n$ converges absolutely if $|x^2| < R$, and diverges if $|x^2| >R$. Hence the radius of convergence must be $\sqrt{R}$.
For the second, note that if we write the power series as $\sum_{n=0}^{\infty}a_n x^n$, then $a_n = 0$ if $n$ is not a square, and $a_n =c_k$ if $n=k^2$. Hence the radius of convergence is $\frac{1}{\rho} = \limsup_n \sqrt[n^2]{|c_n|} $.
We have $\frac{1}{R} = \limsup_n \sqrt[n]{|c_n|}$, hence for all $\epsilon>0$, there exists a subsequence $n_k$ such that $\frac{1}{R} - \epsilon \le \sqrt[n_k]{|c_{n_k}|}$, and for all $\epsilon>0$, there exists $n$ such that $\frac{1}{R} + \epsilon \ge \sqrt[k]{|c_{k}|}$ for all $k \ge n$.
Hence we have $\sqrt[n_k]{\frac{1}{R} - \epsilon} \le \sqrt[n_k^2]{|c_{n_k}|}$, and taking $\limsup$s we have $1 \le \frac{1}{\rho}$. Similarly, the other inequality gives $1 \ge \frac{1}{\rho}$. Hence $\rho = 1$.
It converges conditionally and not absolutely because $\displaystyle \sum_{n=2}^\infty \left|(-1)^n\cdot \dfrac{1}{\ln n}\right| = \displaystyle \sum_{n=2}^\infty \dfrac{1}{\ln n} \geq \displaystyle \sum_{n=2}^\infty \dfrac{1}{n} = \infty$, and you've done the first part that series is convergent by the alternating series test.
Best Answer
For the second limit: $n^2 \leq (n^2+1) \leq (2n^2)$ for $n\geq1$, so we have:
$$n^{\frac{2}{n}} \leq (n^2+1)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}n^{\frac{2}{n}}$$
I'm guessing you have as a standard result that $n^{\frac{1}{n}} \rightarrow 1$ as $n \rightarrow \infty$ and $a^{\frac{1}{n}} \rightarrow 1$ as $n \rightarrow \infty$ for $a > 0$. Then by the product (both sequences on either side of the inequality go to 1) and squeeze rule, $(n^2+1)^{\frac{1}{n}} \rightarrow 1$.
As for the final one:
$$\left(\frac{1}{n^k}\right)^{\frac{1}{n}} = n^{\frac{-k}{n}} = \left(n^{\frac{1}{n}}\right)^{-k}$$
Since $k$ is fixed, and we have $n^{\frac{1}{n}} \rightarrow 1$, by the product/quotient rule, $\left(\frac{1}{n^k}\right)^{\frac{1}{n}} \rightarrow 1$.