= Solution
\b[Yes: use <superdense coding>, with a shared <Bell state>.] Let the sender and receiver initially hold the first and second <qubits> of $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. To encode bits $(a,b)$, the sender applies $Z^aX^b$ to her <qubit>. These <Pauli operators> produce four orthogonal states:
$$
\begin{array}{c|c}
(a,b)&(Z^aX^b\otimes I)|\Phi^+\rangle\\
(0,0)&(|00\rangle+|11\rangle)/\sqrt2\\
(1,0)&(|00\rangle-|11\rangle)/\sqrt2\\
(0,1)&(|01\rangle+|10\rangle)/\sqrt2\\
(1,1)&(|01\rangle-|10\rangle)/\sqrt2
\end{array}
$$
She sends this <qubit> through the noiseless <quantum channel>. The receiver then possesses both <qubits>. He applies a <controlled-NOT gate> from the received <qubit> to his original <qubit>, followed by a <Hadamard gate> on the received <qubit>. This maps the four states in the table to $|ab\rangle$. A measurement in the <computational basis> therefore recovers the two bits with certainty. No classical message is sent.
The resource assumption deserves care: one-qubit <superdense coding> needs a <Bell pair> already shared before encoding. If none is initially available, the sender can prepare a <Bell pair> locally and send one half through the same noiseless <quantum channel> in advance. The protocol then uses two transmitted <qubits> in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two <computational basis> <qubits> directly also communicates two bits. \b[With a pre-shared <Bell pair>, one transmitted <qubit> suffices; without it, the entanglement-distribution cost must be counted.] An unassisted single transmitted <qubit> cannot perfectly encode four classical messages: its <Holevo bound> is at most $\log_2 2=1$ bit.
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