= Solution
Put $q=\cos^2(\pi/8)=(2+\sqrt2)/4$ and $1-q=\sin^2(\pi/8)$. In the <computational basis>, input $|a\rangle$ has outcome probabilities $(q,1-q)$, while $|b\rangle$ has $(1-q,q)$. The marginal outcomes are equally likely; the <Bayesian posterior> probabilities are
$$
\begin{array}{c|c|cc}
\text{outcome}&P(\text{outcome})&P(a\mid\text{outcome})&P(b\mid\text{outcome})\\\hline
0&1/2&q&1-q\\
1&1/2&1-q&q
\end{array}.
$$
Guess $a$ after zero and $b$ after one. Each guess is correct with probability \b[$q\simeq0.853553$] and wrong with probability \b[$1-q\simeq0.146447$]. Neither outcome is certain evidence for one input.
For a real state $|\psi_\theta\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle$, the <Born rule> in the <Hadamard basis> gives $P(\pm\mid\theta)=(1\pm\sin2\theta)/2$. Both angles $\pi/8$ and $3\pi/8$ have $\sin2\theta=1/\sqrt2$. Thus both preparations give $P(+)=q$ and $P(-)=1-q$. By <Bayes' theorem>, \b[either outcome leaves the two input hypotheses equally likely]. The biased outcome frequencies carry no information about which state was prepared; the best success probability with this measurement is \b[$1/2$].
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