= Solution
Set
$$
u=\frac{1-z}{2}.
$$
This Möbius change sends $z=1,-1,\infty$ to $u=0,1,\infty$, respectively, without changing the corresponding exponents. The hypergeometric P-symbol has exponents
$$
\begin{array}{c|ccc}
&0&1&\infty\\ \hline
&0&0&A\\
&1-C&C-A-B&B.
\end{array}
$$
Matching it with $(0,1/2)$ at both finite singularities and $(-n,n)$ at infinity gives
$$
C=\frac12,\qquad A=n,\qquad B=-n.
$$
The exponent-zero solution normalized to one at $u=0$ is therefore
$$
\boxed{w_1(z)=F\left(n,-n;\frac12;\frac{1-z}{2}\right).}
$$
The second standard local solution at $u=0$ is
$$
u^{1-C}F(A-C+1,B-C+1;2-C;u).
$$
After absorbing the constant $2^{-1/2}$ into its normalization, this becomes
$$
\boxed{
w_2(z)=(1-z)^{1/2}
F\left(-n+\frac12,n+\frac12;\frac32;\frac{1-z}{2}\right).}
$$
Their exponents at $z=1$ are $0$ and $1/2$, so they are linearly independent.
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