= Solution
Expand the initial <wavefunction> in the <angular momentum> <eigenbasis> using $\cos^2\phi=1/2+(e^{2i\phi}+e^{-2i\phi})/4$. With the normalized $u_m$ from part (a), this gives
$$
\boxed{|\Psi(0)\rangle=\sqrt{\frac23}|m=0\rangle+\frac1{\sqrt6}|m=2\rangle+\frac1{\sqrt6}|m=-2\rangle.}
$$
The coefficient squares sum to one, also verifying the stated normalization. The <Born rule> therefore gives <angular momentum> outcomes
$$
\boxed{L_z=0,\ +2\hbar,\ -2\hbar\quad\text{with probabilities}\quad\frac23,\ \frac16,\ \frac16.}
$$
For an <energy> measurement the two distinct $m=\pm2$ states belong to the same <energy> eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtain
$$
\boxed{E=0\ \text{with probability }\frac23,\qquad E=\frac{2\hbar^2}{I}\ \text{with probability }\frac13.}
$$
Their relative probabilities are $4:1:1$ for the angular momenta and $2:1$ for the energies. In particular, zero mean <angular momentum> does not imply zero rotational <energy>.
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