Answer
$a.\quad R=1. \quad $Interval of convergence:$\quad -3\lt x\leq-1$
$b.\quad $ Interval of absolute convergence:$\quad -3\lt x\lt-1$
$c.\quad $ At $x=-1,\quad $ the series converges conditionally.
Work Step by Step
(See text: "How to Test a Power Series for Convergence".)
$\text{Step 1.}$
Use the Ratio Test to find the interval where the series converges absolutely.
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(-1)^{n+1}(x+2)^{n+1}}{n+1}\div\frac{(-1)^{n}(x+2)^{n}}{n}\\
&=\displaystyle \frac{(-1)^{n+1}(x+2)^{n+1}}{n+1}\cdot\frac{n}{(-1)^{n}(x+2)^{n}}\\
&=\displaystyle \frac{-(x+2)\cdot n}{n+1}\\
\displaystyle \lim_{n\rightarrow\infty}|-(x+2)\cdot\frac{n}{n+1}|&\lt 1\displaystyle \\
|-(x+2)|\displaystyle \lim_{n\rightarrow\infty}|\frac{n}{n+1}|&\lt 1\displaystyle \\
\text{ (the limit =1)} \\
|x+2|\cdot 1&\lt 1 \\
-1&\lt x+2\lt 1 \\
-3&\lt x\lt-1 \end{align*}.$
Determine the center and radius:
$a-R\lt x\lt a+R$
$-2-1\lt x\lt -2+1$
$a=-2,\quad $
the radius is $R=1,$
the interval of absolute convergence is $-3\lt x\lt-1$
$\text{Step 2.}$
If the interval of absolute convergence is finite, test for convergence or divergence at each endpoint.
$ x=-3\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}(-3+2)^{n}}{n}=\sum_{n=1}^{\infty}\frac{1}{n}\qquad$ ... a divergent series
$ x=-1\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}(-1+2)^{n}}{n}=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\qquad$
... an alternating harmonic series is convergent
$\rightarrow $ conditional convergence at $x=-1$.
$\text{Step 3.}$
If the interval of absolute convergence is $a-R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$a.\quad R=1. \quad $Interval of convergence:$\quad -3\lt x\leq-1$
$b.\quad $ Interval of absolute convergence:$\quad -3\lt x\lt-1$
$c.\quad $ At $x=-1,\quad $ the series converges conditionally.
