= Solution
Diagonalizing the $J_1$ <matrix> gives the normalized <eigenvectors>
$$
v_+=\frac12\begin{pmatrix}1\\\sqrt2\\1\end{pmatrix},\qquad
v_0=\frac1{\sqrt2}\begin{pmatrix}1\\0\\-1\end{pmatrix},\qquad
v_-=\frac12\begin{pmatrix}1\\-\sqrt2\\1\end{pmatrix},
$$
with <eigenvalues> $+\hbar,0,-\hbar$, respectively. The first <projective measurement> prepares $v_+$ up to <global phase>, regardless of the earlier state, conditional on obtaining $+\hbar$.
The <unitary time evolution> under $H=\omega J_3$ gives, with $\theta=\omega t$,
$$
|\Psi(t)\rangle=\operatorname{diag}(e^{-i\theta},1,e^{i\theta})v_+
=\frac12\begin{pmatrix}e^{-i\theta}\\\sqrt2\\e^{i\theta}\end{pmatrix}.
$$
The <probability amplitudes> for the second <projective measurement> are
$$
\langle v_+|\Psi(t)\rangle=\frac{1+\cos\theta}{2},\qquad
\langle v_0|\Psi(t)\rangle=-\frac{i\sin\theta}{\sqrt2},\qquad
\langle v_-|\Psi(t)\rangle=\frac{\cos\theta-1}{2}.
$$
The <Born rule> therefore gives the general <spin-one transverse measurement after axial precession>:
$$
\Pr(+\hbar)=\cos^4\frac\theta2,\qquad
\Pr(0)=\frac12\sin^2\theta,\qquad
\Pr(-\hbar)=\sin^4\frac\theta2.
$$
At the specified times $t_k=\pi k/\omega$, $\theta=\pi k$, so
$$
\boxed{\Pr(+\hbar)=\frac{1+(-1)^k}{2},\qquad
\Pr(0)=0,\qquad
\Pr(-\hbar)=\frac{1-(-1)^k}{2}.}
$$
For even $k$, the result is certainly $+\hbar$; for odd $k$, it is certainly $-\hbar$. Although zero belongs to the spectrum of the <observable>, it never occurs at these times. Indeed, for even $k$ the evolved vector is $v_+$, while for odd $k$ it is $-v_-$, where the minus sign is a <global phase>.
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