= Solution
Interpret the imposed measurement restriction as direct measurement of the encoded two-qubit <Bell state> in the <computational basis>, without a preceding entangling Bell decoder. For the uniform messages the conditional probabilities are
$$
\begin{array}{c|cccc}
&00&01&10&11\\\hline
\Psi^+&0&1/2&1/2&0\\
\Psi^-&0&1/2&1/2&0\\
\Phi^+&1/2&0&0&1/2\\
\Phi^-&1/2&0&0&1/2
\end{array}.
$$
Each outcome has probability $1/4$, so $H(Y)=2$ bits, while $H(Y|X)=1$ bit. Therefore
$$
\boxed{I(X:Y)=1\ \text{bit}.}
$$
The outcome reveals even versus odd parity, but none of the plus/minus label. Each outcome is compatible with two equally probable messages.
Now insert an arbitrary single-qubit <unitary operator> $U$ on one wire immediately before measurement. Measuring that wire is equivalent to measuring along a <Bloch vector> $\boldsymbol n$, with the other wire still measured along $z$. All four <Bell states> have zero local means, and $\langle X\otimes Z\rangle=\langle Y\otimes Z\rangle=0$, whereas $\langle Z\otimes Z\rangle=+1$ for $\Phi^\pm$ and $-1$ for $\Psi^\pm$. Hence
$$
\Pr(r,s\mid\Phi^\pm)=\tfrac14(1+rsn_z),\qquad
\Pr(r,s\mid\Psi^\pm)=\tfrac14(1-rsn_z),\qquad r,s=\pm1.
$$
The sign within each pair is still inaccessible, and
$$
\boxed{I_U(X:Y)=1-h_2\left(\frac{1+n_z}{2}\right)\leq1.}
$$
It stays one for a computational-axis measurement, but a <Hadamard gate> makes $n_z=0$ and reduces it to zero. Thus a single local gate can change the information, by reducing it, and cannot increase it above one bit.
There is an important scope qualification: if Bob is allowed the full two-qubit decoder from part b before his final computational measurement, those measurements already identify all four messages and give two bits. The one-bit result refers to restricting the Bell readout itself to product computational measurements. Merely specifying the final detector basis without restricting the decoder would not impose that one-bit limit.
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