= Solution
Put $r_s=2\mu$ and $f(r)=1-r_s/r$. Radial <null geodesics> of the <Schwarzschild metric> obey
$$
\frac{dr}{dt}=\pm c f(r),\qquad
ct=\pm\left[r+r_s\log\left|\frac r{r_s}-1\right|\right]+\text{constant}.
$$
Near the horizon, $\delta=r-r_s$ satisfies
$$
\boxed{\delta(t)\simeq C e^{\pm ct/r_s}.}
$$
$C>0$ describes the exterior and $C<0$ the interior. The exponential approach in Schwarzschild coordinate time resembles part (a), but $\delta$ is not a proper radial distance.
Outside, set $\rho=2\sqrt{r_s\delta}$ and retain the leading near-horizon terms. The radial metric becomes the <Rindler limit of a Schwarzschild horizon>:
$$
ds^2\simeq\left(\frac{c\rho}{2r_s}\right)^2dt^2-d\rho^2.
$$
It agrees with part (a) for $x=\rho$, $\alpha=c/(2r_s)$. Accordingly $\rho\propto e^{\pm ct/(2r_s)}$, while $\delta\propto\rho^2$ explains the factor two in its exponent.
Inside, write $r=r_s-\rho^2/(4r_s)$. The leading radial metric is instead
$$
ds^2\simeq d\rho^2-\left(\frac{c\rho}{2r_s}\right)^2dt^2.
$$
Here the radial coordinate is timelike and the Schwarzschild $t$ coordinate is spacelike. Thus $x<0$ in the Rindler metric does not represent the black-hole interior: $g_{tt}=(\alpha x)^2$ stays positive on both Rindler wedges. In the black-hole interior all future-directed radial light rays move towards decreasing $r$; the two signs of $dr/dt$ do not mean opposite future radial directions because $t$ is no longer time.
The horizon is a <coordinate singularity> in Schwarzschild coordinates and in the accelerated chart, not a curvature singularity. The full Schwarzschild geometry nevertheless has genuine <tidal forces>; the flat Rindler metric captures its leading local radial near-horizon structure.
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