(a)
By Ratio Test:
$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{2^{n+1} z^{n+1} {n+1}}{3^{n+1}} \frac{3^{n}}{2^{n} z^{n} n}| \\
=\lim_{n \to \infty} |\frac{n+1}{n} \frac{2}{3} z| \\
= |\frac{2}{3} z|
$
We want $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1$:
$
|\frac{2}{3} z| < 1 \\
|z| < \frac{3}{2} \\
R = \frac{3}{2}
$
Now testing $|z| = \frac{3}{2}$:
$
\sum\limits_{n=1}^{\infty} |\frac{2^{n} z^{n} {n}}{3^{n}}| \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |z|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |\frac{3}{2}|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} |n|
$
The series diverges at the boundary, thus we have:
$|z| < \frac{3}{2}$
(b)
By Ratio Test:
$
\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{ {(n+1)}! z^{n+1} }{{(2n + 2)}!} \frac{{(2n)}!}{{n}! z^{n}}| \\
=\lim_{n \to \infty} |\frac{n+1}{(2n+1)(2n+2)}z| \\
= 0
$
Thus, $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1 \forall z$
Therefore, we have the series converges for:
$
R = \infty \\
|z| < \infty
$
