= Solution
With $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$, the matrix of $S_z\cos\theta+S_x\sin\theta$ is $(\hbar/2)\begin{pmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{pmatrix}$. Multiplication by $(c,s)^T$ and $(-s,c)^T$ gives <eigenvalues> $+\hbar/2$ and $-\hbar/2$, using the double-angle identities.
The one-particle transformation is the matrix $R=\begin{pmatrix}c&-s\\s&c\end{pmatrix}$, with <determinant> one. The antisymmetric tensor $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ transforms by this <determinant> under $R\otimes R$, so the singlet is unchanged. The total $S_z$ kills it. Its invariance under these rotations implies that total $S_z\cos\theta+S_x\sin\theta$ also kills it for every $\theta$, so total $S_x$ kills it. The spin commutator then gives total $S_y$ acting as zero. Therefore \b[$S^2|\chi\rangle=0$] and it has total spin zero.
Rotating only particle B produces
$$
|\chi_\theta\rangle=\frac1{\sqrt2}(-s|\uparrow\uparrow\rangle+c|\uparrow\downarrow\rangle-c|\downarrow\uparrow\rangle-s|\downarrow\downarrow\rangle).
$$
The measurement <probabilities> are therefore
$$
\boxed{P(\uparrow,\uparrow)=P(\downarrow,\downarrow)=\tfrac12\sin^2(\theta/2),\qquad P(\uparrow,\downarrow)=P(\downarrow,\uparrow)=\tfrac12\cos^2(\theta/2).}
$$
Each arrow represents the corresponding value $\pm\hbar/2$. The sequential measurements commute because they act on different particles. The two parallel outcomes leave <triplet states> with total spin one and \b[$S^2=2\hbar^2$]. Each antiparallel <product state> is a superposition of the spin-zero singlet and the spin-one zero-projection triplet, and is not an $S^2$ eigenstate. Outcomes of zero <probability> at a special angle do not actually occur.
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