To answer your first question about the boundaries of your region of convergence: Let us try putting $x^2=1/3$ ($x$ is on the boundary), and see if the series converges. The series is now $$\sum_{n=0}^{\infty}\frac{(-1)^n}{\sqrt{n^2+4}}.$$ According to the alternating series test, this series converges because $\lim_{n\to \infty}\frac{1}{\sqrt{n^2+4}}=0,$ which means that the boundary IS part of the region of convergence: $$x^2\leq1/3.$$
The quotient test in the second question gives $$\lim_{n\to\infty}\big|\frac{a_{n+1}}{a_n}\big|=\lim_{n\to\infty}|x+1|\cdot\frac{4^n+n^5}{4^{n+1}+(n+1)^5}.$$
To find the limit of the fraction, we can simply notice that the exponential functions will dominate for large $n$, so that $$\lim_{n\to \infty}\frac{4^n+n^5}{4^{n+1}+(n+1)^5}=\lim_{n\to \infty} \frac{4^n}{4^{n+1}}=1/4.$$
Alternatively, we can apply l'Hôpital's theorem repeatedly. Applying it once gives:
$$\lim_{n\to \infty}\frac{4^n+n^5}{4^{n+1}+(n+1)^5}=\lim_{n\to \infty} \frac{\ln{4}\cdot4^n+5n^4}{\ln{4}\cdot4^{n+1}+5(n+1)^4}.$$
Doing this 5 more times will kill off the polynomial term, and results in
$$\lim_{n\to \infty} \frac{(\ln{4})^6\cdot4^n}{(\ln{4})^6\cdot4^{n+1}}=1/4.$$
The quotient test then reads
$$\lim_{n\to\infty}\big|\frac{a_{n+1}}{a_n}\big|=\frac{1}{4}|x+1|,$$
and so the series converges for $$|x+1|<4$$ with a radius of convergence of 4. What about the boundry? Inserting $x+1=\pm4$ in the original series: $$\sum_{n=0}^{\infty}\pm(-1)^n\cdot\frac{4^{n+3}}{4^n+n^5}.$$ This series does NOT converge according to the alternating series theorem because $$\lim_{n\to \infty}\frac{4^{n+3}}{4^n+n^5}=4^3\neq0,$$ and so the boundary is not a part of the region of convergence.
