= Solution
Use the <Hadamard gate> identity $H=(X+Z)/\sqrt2$ on the first data wire. The $X_1$ branch is corrected as above and has syndrome $01$. The $Z_1$ branch changes the encoded state to $\alpha|000\rangle-\beta|111\rangle$, with syndrome $00$. No bit-flip recovery is triggered, and decoding leaves $Z|\psi\rangle$ on the top wire.
By linearity of the full <quantum circuit>, its output for $H_1$ is therefore
$$
\frac1{\sqrt2}\left(|\psi\rangle\otimes|0001\rangle+Z|\psi\rangle\otimes|0000\rangle\right),
$$
where the four-bit strings are ordered as wires $2,3,4,5$. The two syndrome states are orthogonal. Taking their <partial trace> removes the cross terms, so the <reduced density matrix> is
$$
\rho=\frac12\left(|\psi\rangle\langle\psi|+Z|\psi\rangle\langle\psi|Z\right)
=\boxed{\begin{pmatrix}|\alpha|^2&0\\0&|\beta|^2\end{pmatrix}.}
$$
The top output undergoes a <completely dephasing channel> in the computational basis. The bit-flip component is corrected, but the undetected phase component destroys logical coherence after the syndrome is ignored.
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