For question a, we have
\begin{equation}
f(z)=(1-z)^{-1/2} \\
a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{f^{(n)}(0)}{n!}
\end{equation}
Then the $nth$ derivative of $f(z)$ can be derived as
\begin{equation}
f^\prime(z) = \frac{1}{2}(1-z)^{-3/2} \\
f''(z) = \frac{3}{4}(1-z)^{-5/2} \\
f'''(z) = \frac{15}{8} \times (1-z)^{-7/2} \\
f''''(z) =\frac{105}{16} \times (1-z)^{-9/2}
\end{equation}
At $z=0$
\begin{equation}
f(0) = 1
f'(0) = \frac{1}{2} \\
f''(0) = \frac{3}{4} \\
f'''(0) = \frac{15}{8} \\
f''''(0) = \frac{105}{16} \\
f^{(n)}(0) = \frac{1 \times 3 \times \dots \times (2n-1)}{2^n} \\
a_n = \frac{1 \times 3 \times 5 \times \dots \times (2n-1)}{2^n n!}
\end{equation}
Thus we have the power series
\begin{equation}
f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 1 + \frac{z}{2} + \frac{3z^2}{8}+ \dots
\end{equation}
The radius of convergence is
\begin{equation}
\frac{1}{R} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{f^{(n+1)}(0)}{(n+1)!} \times \frac{n!}{f^{(n)}(0)} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2n+2} \mid = 1 \\
R = 1
\end{equation}
For question b, let
\begin{equation}
F(z) = \arcsin(z) \\
F'(z)=\frac{1}{\sqrt{1-z^2}} = (1-z^2)^{-1/2}
\end{equation}
Note that
\begin{equation}
f(z^2)=(1-z^2)^{-1/2} \Rightarrow F'(z)=f(z^2) \\
F(z) = \int f(z^2)
\end{equation}
Then
\begin{equation}
f(z^2) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n!}z^2n \\
F(z) = \int f(z^2) = \int \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n!}z^{2n} dz
F(z) = (\sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n! \cdot (2n+1)}z^{2n+1}) +C
\end{equation}
Since $F(0) = 0 \Rightarrow C=0$
\begin{equation}
F(z) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n! \cdot (2n+!)}z^{2n+1}
\end{equation}
