= Solution
In $c=k_B=\hbar=1$ units, the <Hawking temperature> is $T_H=|\kappa|/(2\pi)$ for the stated time normalization. The inertial central observer sees the <de Sitter horizon temperature> $T_0=1/(2\pi R)$. The <Tolman temperature law> states $T\sqrt{-k^2}=\text{constant}$ in static <thermal equilibrium>. \b[Hence]
$$
\boxed{T(r)=\frac1{2\pi R\sqrt{1-r^2/R^2}}.}
$$
With part b, $T^2=(a^2+R^{-2})/(4\pi^2)$ and $T/(a/(2\pi))=R/r\to1$ at the horizon. This tends to the <Unruh effect> temperature for the increasingly accelerated observer.
Let $s$ be proper distance inward from the horizon. Then $R-r\simeq s^2/(2R)$, $F\simeq s^2/R^2$, giving
$$
ds^2\simeq-\frac{s^2}{R^2}dt^2+ds^2+R^2d\Omega_2^2.
$$
The radial factor has <Rindler coordinates>: $T_{\rm M}=s\sinh(t/R)$ and $X_{\rm M}=s\cosh(t/R)$ make it Minkowskian. Fixed $s$ has acceleration $1/s$ and temperature $1/(2\pi s)$, agreeing with the leading behavior. At the center $a=0$ but the finite de Sitter temperature remains; $T=a/(2\pi)$ is only the near-horizon limit here.
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