Answer
$a.\quad R=1. \quad $Interval of convergence:$\quad -1\lt x\lt 1$
$b.\quad $ Interval of absolute convergence:$\quad -1\lt x\lt 1$
$c.\quad $ No values of x for which 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.
Ordinarily, $|x-a|\lt R\quad $ or $\quad a-R\lt x\lt a+R.$
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(n+1)x^{n+1}}{(n+3)}\div\frac{nx^{n}}{(n+2)}\\
&=\displaystyle \frac{(n+1)x^{n+1}}{(n+3)}\div\frac{n+2}{nx^{n}}\\
&=\displaystyle \frac{x(n+1)(n+2)}{n(n+3)}\\
\displaystyle \lim_{n\rightarrow\infty}|\frac{x(n+1)(n+2)}{n(n+3)}|&\lt 1\displaystyle \\
|x|\displaystyle \lim_{n\rightarrow\infty}|\frac{(n+1)(n+2)}{n(n+3)}|&\lt 1\displaystyle \\
\end{align*}.$
We see that both polynomials have degree 2. The leading coefficients are the same, so the limit equals 1.
$\begin{align*}
|x|\cdot 1&\lt 1\\
-1&\lt x\lt 1\\ \end{align*}$
Determine the center and radius:
$a-R\lt x\lt a+R$
$0-1\lt x\lt 0+1$
$a=0,\quad $
the radius is $R=1,$
the interval of absolute convergence is $-1\lt x\lt 1$
$\text{Step 2.}$
If the interval of absolute convergence is finite,
test for convergence or divergence at each endpoint.
$ x=-1\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+2}\qquad$
$\displaystyle \lim_{n\rightarrow\infty}\frac{(-1)^{n}n}{n+2} $ ... is not zero (it fails to exist, alternates to $\pm 1$)
... by the n-th term test, this is a divergent series
$ x=1\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+2}\qquad$
... divergent series (the limit of the n-th term is 1, not 0)
$\text{Step 3.}$
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 -1\lt x\lt 1$
$b.\quad $ Interval of absolute convergence:$\quad -1\lt x\lt 1$
$c.\quad $ No values of x for which the series converges conditionally.