= Solution
Using the states from part (a),
$$
|\chi\rangle=\frac1{\sqrt2}
\left[
-s|\uparrow\uparrow\rangle
+c|\uparrow\downarrow\rangle
-c|\downarrow\uparrow\rangle
-s|\downarrow\downarrow\rangle
\right].
$$
The <Born rule> therefore gives
$$
\begin{array}{c|c}
\text{outcome}&\text{probability}\\ \hline
(+\hbar/2,+\hbar/2)&\frac12\sin^2(\theta/2)\\
(+\hbar/2,-\hbar/2)&\frac12\cos^2(\theta/2)\\
(-\hbar/2,+\hbar/2)&\frac12\cos^2(\theta/2)\\
(-\hbar/2,-\hbar/2)&\frac12\sin^2(\theta/2).
\end{array}
$$
After the equal-sign outcomes the state is respectively $|\uparrow\uparrow\rangle$ or $|\downarrow\downarrow\rangle$. These are <spin-one-half triplet state>[triplet states], hence <eigenstates> of $\mathbf S^2$ with eigenvalue
$$
\boxed{2\hbar^2}.
$$
The opposite-sign product states are each a nontrivial <linear combination> of the singlet and the $m=0$ triplet, so they are not eigenstates of $\mathbf S^2$.
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