= Solution
Write $\omega=\gamma B$. Since
$$
H=-\omega S_z,
$$
the time-evolution operator is
$$
U(t)=e^{i\omega tS_z/\hbar}.
$$
For the initial state $|\hat{\mathbf z},3/2\rangle=|3/2\rangle$,
$$
\boxed{|\psi(t)\rangle=e^{3i\omega t/2}|3/2\rangle.}
$$
Only an overall phase changes, so every spin measurement has time-independent probabilities.
For the initial state along $\hat{\mathbf x}$, part (c) with $\varphi=0$ gives
$$
\boxed{
|\psi(t)\rangle=\frac1{2\sqrt2}
\begin{pmatrix}
e^{3i\omega t/2}\\
\sqrt3e^{i\omega t/2}\\
\sqrt3e^{-i\omega t/2}\\
e^{-3i\omega t/2}
\end{pmatrix}.}
$$
Up to the overall phase $e^{3i\omega t/2}$, this is the coherent state of part (c) with $\varphi=-\omega t$. Thus the spin direction undergoes Larmor precession about the $z$-axis with angular velocity $-\gamma B$.
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