= Solution
<matrix> multiplication gives $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$, hence $[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$. For physical <neutron> <spin>, use $\mathbf S=(\hbar/2)\boldsymbol\sigma$, so $[S_i,S_j]=i\hbar\epsilon_{ijk}S_k$, $S^2=3\hbar^2/4$, and measurement along any axis gives \b[$\pm\hbar/2$].
For the specified field, $H=(\alpha B\hbar/2)\sigma_y$. The <Time-dependent Schrödinger equation> is $i\hbar\dot\psi=H\psi$ with $\psi(0)=(1,0)^T$. The <uniform-field spin precession> follows from $\sigma_y^2=I$:
$$
\psi(t)=e^{-i\alpha Bt\sigma_y/2}\binom10
=\boxed{\binom{\cos\phi}{\sin\phi}},\qquad
\boxed{\phi(t)=\alpha Bt/2}.
$$
The sign of the precession is contained in the given $\alpha$. The normalized $x$-axis eigenstates are \b[$(1,1)^T/\sqrt2$ and $(1,-1)^T/\sqrt2$].
The measurement direction is $\mathbf n=(0,\sin\theta,\cos\theta)$, as confirmed by the original diagram. Write $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. The spinors $\chi_+=(c,is)^T$ and $\chi_-=(is,c)^T$ have unit norm and <inner product> $ics-isc=0$. Applying $\mathbf n\cdot\boldsymbol\sigma$ gives <eigenvalues> $+1$ and $-1$. At $\theta=0$ they are the usual $z$ basis; at $\pi/2$ they are the $y$ basis up to a <phase>; at $\pi$ the $z$ up/down roles are exchanged, with a factor $i$; and at $2\pi$ both return with a minus sign. A spinor changes sign under a full rotation, while its isolated-state measurement <probabilities> are unchanged.
For <spin measurement in the yz plane>, projection gives
$$
P_+=|c\cos\phi-is\sin\phi|^2
=\boxed{\frac{1+\cos\theta\cos(\alpha Bt)}2},\qquad
\boxed{P_-=\frac{1-\cos\theta\cos(\alpha Bt)}2}.
$$
For $\theta=0$ these are $\cos^2\phi,\sin^2\phi$; for $\pi/2$ both are $1/2$ at all times; for $\pi$ they are exchanged; and for $2\pi$ they are the same as at zero angle. The constant $y$ <probabilities> also follow from conservation of $S_y$.
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