= Solution
Direct <matrix multiplication> gives
$$
\begin{array}{c|cc}
&|\uparrow\rangle&|\downarrow\rangle\\\hline
\hat S_x&\frac\hbar2|\downarrow\rangle&\frac\hbar2|\uparrow\rangle\\
\hat S_y&\frac{i\hbar}2|\downarrow\rangle&-\frac{i\hbar}2|\uparrow\rangle\\
\hat S_z&\frac\hbar2|\uparrow\rangle&-\frac\hbar2|\downarrow\rangle
\end{array}
$$
The normalized $S_x$ <eigenvectors> are $(|\uparrow\rangle\pm|\downarrow\rangle)/\sqrt2$, and the $S_y$ <eigenvectors> are $(|\uparrow\rangle\pm i|\downarrow\rangle)/\sqrt2$, with <eigenvalues> $\pm\hbar/2$ in each case.
For $N=|a|^2+|b|^2>0$, allowing for a possibly unnormalized input pair of coefficients,
$$
\boxed{\langle\hat S_z\rangle=\frac\hbar2\frac{|a|^2-|b|^2}{N},\qquad\langle\hat S^2\rangle=\frac{3\hbar^2}{4}.}
$$
For a normalized state $N=1$. The second result follows because each squared <Spin operator> matrix equals $\hbar^2\hat I/4$.
The tilted component is $\hat S_{z'}=(\hbar/2)\begin{pmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{pmatrix}$. Its <eigenvalues> remain $\pm\hbar/2$, with one real phase convention for the normalized <eigenvectors>
$$
\boxed{|+z'\rangle=\cos(\theta/2)|\uparrow\rangle+\sin(\theta/2)|\downarrow\rangle,\qquad|-z'\rangle=-\sin(\theta/2)|\uparrow\rangle+\cos(\theta/2)|\downarrow\rangle.}
$$
Projecting the initial $|\uparrow\rangle$ onto them gives \b[$P_+=\cos^2(\theta/2)$] and \b[$P_-=\sin^2(\theta/2)$]. Thus
$$
\boxed{\langle\hat S_{z'}\rangle=\frac\hbar2(P_+-P_-)=\frac\hbar2\cos\theta.}
$$
A smaller <expectation> does not imply a smaller individual <spin> <eigenvalue>; it describes the probabilities of the same two measurement outcomes.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2016/paper-1-spin-projection.png]
{title=Rotating the measurement axis changes spin-half outcome probabilities rather than the outcome magnitudes}
{source=paper-1-spin-projection.py}
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