= Solution
Keep one normalized <spin> function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta,\phi)=\sin\theta\,e^{im\phi}$, with $m=\pm1$. Overall constants and the <spin> factor pass unchanged through the spatial operators.
First, in <atomic units>, $\hat l_z=-i\partial_\phi$, so
$$
\hat l_z\Psi=-i(im)\Psi=\boxed{m\Psi}.
$$
For the squared <orbital angular momentum>, $\partial_\phi^2f_m=-f_m$ and
$$
\partial_\theta\!\left(\sin\theta\,\partial_\theta f_m\right)
=(\cos^2\theta-\sin^2\theta)e^{im\phi}.
$$
Therefore the two terms of the angular operator give
$$
\hat l^2f_m=
\frac{-\cos^2\theta+\sin^2\theta+1}{\sin\theta}e^{im\phi}
=2\sin\theta e^{im\phi}=2f_m,
$$
using $1-\cos^2\theta=\sin^2\theta$. Thus \b[$\hat l^2\Psi=2\Psi$].
Finally, the spatial hydrogen <Hamiltonian> is
$$
\hat H=-\frac12\left[\frac{1}{r^2}\partial_r(r^2\partial_r)-\frac{\hat l^2}{r^2}\right]-\frac1r.
$$
For $R=Ar e^{-r/2}$,
$$
R'=A(1-r/2)e^{-r/2},\qquad
\frac1{r^2}\frac{\mathrm d}{\mathrm dr}(r^2R')
=Ae^{-r/2}\left(\frac2r-2+\frac r4\right).
$$
Substitute the already established angular <eigenvalue> $l(l+1)=2$:
$$
\hat H(RY\sigma)=AY\sigma e^{-r/2}
\left[-\frac12\left(\frac2r-2+\frac r4\right)+\frac1r-1\right]
=-\frac18RY\sigma.
$$
Hence \b[$E=-1/8$ hartree], consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.
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