= Solution
Put $f_l(\theta,\phi)=\sin^l\theta\,e^{-il\phi}$. For the regular, single-valued <orbital angular momentum> states, $l$ is a nonnegative integer. Differentiation immediately gives
$$
L_3f_l=-i\hbar\partial_\phi f_l=(-i\hbar)(-il)f_l=\boxed{-l\hbar f_l}.
$$
For $0<\theta<\pi$, the derivatives needed for $L^2$ are
$$
\partial_\theta f_l=l\cot\theta f_l,\qquad \partial_\theta^2f_l=(l^2\cot^2\theta-l\csc^2\theta)f_l,\qquad \partial_\phi^2f_l=-l^2f_l.
$$
Consequently the angular differential expression inside the parentheses of $-L^2/\hbar^2$ equals
$$
\left[(l^2+l)\cot^2\theta-(l+l^2)\csc^2\theta\right]f_l=-l(l+1)f_l,
$$
using $\csc^2\theta-\cot^2\theta=1$. Therefore
$$
\boxed{L^2f_l=l(l+1)\hbar^2f_l.}
$$
The expressions extend regularly through the polar coordinate singularities: $f_l$ is the restriction of the polynomial $(x-iy)^l$ to the unit sphere. It is proportional to the lowest-projection <spherical harmonic> of degree $l$; for $l=0$ it is the constant state.
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