= Solution
Let $\lambda=e^{2\pi i\phi}$. In terms of the <Pauli X gate>, the given <unitary operator> is $U_\phi=aI+bX$, with $a=(1+\lambda)/2$ and $b=(1-\lambda)/2$. Since $X|+\rangle=|+\rangle$ and $X|-\rangle=-|-\rangle$,
$$
\boxed{U_\phi|+\rangle=|+\rangle,\qquad U_\phi|-\rangle=e^{2\pi i\phi}|-\rangle.}
$$
Thus the normalized <eigenvectors> are $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$, with <eigenvalues> $1$ and $e^{2\pi i\phi}$ respectively. At $\phi=0$ they coincide as eigenvalues, and the whole two-dimensional <Hilbert space> is the eigenspace; the same chosen eigenbasis remains valid.
Prepare $|-\rangle$ independently of $\phi$ and use the <exact quantum phase estimation> circuit of the preceding parts on the controlled-$U_\phi$ black box. It returns $\phi$ exactly using
$$
\boxed{2^m-1=O(2^m)\text{ controlled black-box calls}.}
$$
The chosen eigenstate contains all the unknown phase information, unlike $|+\rangle$, whose eigenvalue is independent of $\phi$.
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