= Solution
Expanding with the <angular momentum> <commutators> gives
$$
\begin{aligned}
[\hat L_z,\hat L_\pm]&=[\hat L_z,\hat L_x]\pm i[\hat L_z,\hat L_y]=\pm\hbar\hat L_\pm,\\
\hat L_-\hat L_+&=(\hat L_x-i\hat L_y)(\hat L_x+i\hat L_y)=\hat L_x^2+\hat L_y^2+i[\hat L_x,\hat L_y]=\hat L^2-\hat L_z^2-\hbar\hat L_z.
\end{aligned}
$$
For $\hat L_z|\psi\rangle=\alpha\hbar|\psi\rangle$, commuting $\hat L_z$ past a <ladder operator> gives
$$
\hat L_z\hat L_\pm|\psi\rangle=(\alpha\pm1)\hbar\hat L_\pm|\psi\rangle.
$$
Thus the raised or lowered vector, \b[if nonzero], is an <eigenvector> with the shifted <eigenvalue>; an <angular momentum ladder endpoint> can instead be annihilated. Also $[\hat L^2,\hat L_\pm]=0$, so $\hat L^2\hat L_\pm|\psi\rangle=\Lambda\hbar^2\hat L_\pm|\psi\rangle$ for every nonzero ladder result.
For <orbital angular momentum>, $\hat L_z=-i\hbar\partial_\phi$ gives an azimuthal factor $e^{i\alpha\phi}$. Single-valuedness under $\phi\mapsto\phi+2\pi$ requires \b[integer $\alpha$]. Positivity of ladder norms gives
$$
\|\hat L_+|\Lambda,m\rangle\|^2=\hbar^2[\Lambda-m(m+1)],\qquad
\|\hat L_-|\Lambda,m\rangle\|^2=\hbar^2[\Lambda-m(m-1)]
$$
for normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda=\ell(\ell+1)$; the lower endpoint is $m=-\ell$. Consequently
$$
\boxed{\hat L^2|\ell,m\rangle=\ell(\ell+1)\hbar^2|\ell,m\rangle,\quad \hat L_z|\ell,m\rangle=m\hbar|\ell,m\rangle,\quad m=-\ell,-\ell+1,\ldots,\ell,}
$$
with $\ell=0,1,2,\ldots$. This integer conclusion is for <orbital angular momentum>; <spin> representations can also have half-integer labels.
For the specified Gaussian-radial state, use a <Cartesian p-orbital decomposition>: write $x+y+z=r(\sin\theta\cos\phi+\sin\theta\sin\phi+\cos\theta)$. Its angular factor is
$$
\sqrt{\frac{2\pi}{3}}\left[(1+i)Y_{1,-1}+\sqrt2Y_{1,0}+(-1+i)Y_{1,1}\right].
$$
Only $\ell=1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome is
$$
\boxed{P(L^2=2\hbar^2,L_z=\hbar)=\frac13.}
$$
The common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.
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