(1)

\begin{align*} \sum_{k=0}^{n}C\left(2k,k\right)z^{k} & =\sum_{k=0}^{n}\frac{\left(2k\right)!}{k!k!}z^{k}\\ & =\sum_{k=0}^{n}\frac{\left(2k\right)!!\left(2k-1\right)!!}{k!k!}z^{k}\\ & =\sum_{k=0}^{n}\frac{2^{k}k!2^{k}\left(k-\frac{1}{2}\right)!}{k!k!\Gamma\left(\frac{1}{2}\right)}z^{k}\\ & =\sum_{k=0}^{n}\frac{4^{k}Q\left(\frac{1}{2},k\right)}{k!}z^{k}\\ & =\sum_{k=0}^{n}\frac{Q\left(\frac{1}{2},k\right)}{k!}\left(4z\right)^{k}\\ & =\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2},k\right)}{k!}\left(4z\right)^{k}-\sum_{k=n+1}^{\infty}\frac{Q\left(\frac{1}{2},k\right)}{k!}\left(4z\right)^{k}\\ & =\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}P\left(-\frac{1}{2},k\right)}{k!}\left(4z\right)^{k}-\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2},n+k+1\right)}{\left(n+k+1\right)!}\left(4z\right)^{n+k+1}\\ & =\sum_{k=0}^{\infty}C\left(-\frac{1}{2},k\right)\left(-4z\right)^{k}-\sum_{k=0}^{\infty}\frac{Q\left(\frac{1}{2},n+1\right)Q\left(\frac{1}{2}+n+1,k\right)}{Q\left(1,n+1\right)Q\left(1+n+1,k\right)}\left(4z\right)^{n+k+1}\\ & =\left(1-4z\right)^{-\frac{1}{2}}-\frac{Q\left(\frac{1}{2},n+1\right)}{Q\left(1,n+1\right)}\left(4z\right)^{n+1}\sum_{k=0}^{\infty}\frac{Q\left(1,k\right)Q\left(n+\frac{3}{2},k\right)}{Q\left(n+2,k\right)}\frac{\left(4z\right)^{k}}{k!}\\ & =\left(1-4z\right)^{-\frac{1}{2}}-\frac{\left(2n+1\right)!}{2^{2n+1}n!\left(n+1\right)!}\left(4z\right)^{n+1}F\left(1,n+\frac{3}{2};n+2;4z\right)\\ & =\left(1-4z\right)^{-\frac{1}{2}}-\frac{2\left(2n+1\right)!}{n!\left(n+1\right)!}z^{n+1}F\left(1,n+\frac{3}{2};n+2;4z\right)\\ & =\left(1-4x\right)^{-\frac{1}{2}}-\frac{\left(2\left(n+1\right)\right)!}{\left(n+1\right)!\left(n+1\right)!}x^{n+1}F\left(1,n+\frac{3}{2};n+2;4x\right)\\ & =\left(1-4z\right)^{-\frac{1}{2}}-C\left(2\left(n+1\right),n+1\right)z^{n+1}F\left(1,n+\frac{3}{2};n+2;4z\right) \end{align*}

(2)

\begin{align*} \sum_{k=0}^{\infty}C\left(2k,k\right)z^{k} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}C\left(2k,k\right)z^{k}\\ & =\lim_{n\rightarrow\infty}\left\{ \left(1-4z\right)^{-\frac{1}{2}}-C\left(2\left(n+1\right),n+1\right)z^{n+1}F\left(1,n+\frac{3}{2};n+2;4z\right)\right\} \\ & =\left(1-4z\right)^{-\frac{1}{2}} \end{align*}