Answer
$a.\displaystyle \quad R=\frac{1}{2}. \quad $Interval of convergence:$\displaystyle \quad -\frac{1}{2}\lt x\lt\frac{1}{2}$
$b.\quad $ Interval of absolute convergence:$\displaystyle \quad -\frac{1}{2}\lt x\lt\frac{1}{2}$
$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 $\quad a-R\lt x\lt a+R.$
$\begin{align*}
\displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(2x)^{n+1}}{(2x)^{n}}\\
&=2x\\
\displaystyle \lim_{n\rightarrow\infty}|2x|&\lt 1\displaystyle \\ \text{ We solve the inequality:} \\
|2x|&\lt 1\\
-1&\lt 2x\lt 1\\
-\displaystyle \frac{1}{2}&\lt x\displaystyle \lt\frac{1}{2} \end{align*}$
Determine the center and radius:
$a-R\lt x\lt a+R$
$0-\displaystyle \frac{1}{2}\lt x\lt 0+\frac{1}{2}$
$a=0,\quad $
the radius is $R=\displaystyle \frac{1}{2},$
the interval of absolute convergence is $-\displaystyle \frac{1}{2}\lt x\lt\frac{1}{2}$
$\text{Step 2.}$
If the interval of absolute convergence is finite,
test for convergence or divergence at each endpoint.
$ x=-\displaystyle \frac{1}{2}\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\qquad$ ... divergent series
$ x=\displaystyle \frac{1}{2}\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}(1)^{n}\qquad$ ... divergent series
$\text{Step 3.}$
If the interval of absolute convergence is $a-R\lt x\lt a+R$,
the series diverges for $|x-a|\gt R.$
So,
$a.\displaystyle \quad R=\frac{1}{2}. \quad $Interval of convergence:$\displaystyle \quad -\frac{1}{2}\lt x\lt\frac{1}{2}$
$b.\quad $ Interval of absolute convergence:$\displaystyle \quad -\frac{1}{2}\lt x\lt\frac{1}{2}$
$c.\quad $ No values of x for which the series converges conditionally.