= Solution
The states $|2\rangle$ and $|3\rangle$ are degenerate and couple to $|1\rangle$ with amplitudes $\varepsilon V_0$ and $\varepsilon^2V_0$. The <bright and dark state reduction of a star-coupled Hamiltonian> gives one exact dark eigenstate proportional to $-\varepsilon|2\rangle+|3\rangle$, with energy $E$. The orthogonal bright state couples to $|1\rangle$ with magnitude
$$
q=V_0\sqrt{\varepsilon^2+\varepsilon^4}
=\varepsilon V_0\sqrt{1+\varepsilon^2}.
$$
Diagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrum
$$
\boxed{
E,\qquad
E_\pm=\frac{E_1+E}{2}
\pm\sqrt{\frac{(E_1-E)^2}{4}
+\varepsilon^2V_0^2(1+\varepsilon^2)}.}
$$
Writing $\Delta=E_1-E\ne0$ and expanding the branches according to the eigenvalue to which they converge,
$$
\begin{aligned}
E_{\text{near }E_1}
&=E_1+\frac{\varepsilon^2V_0^2}{E_1-E}
+O(\varepsilon^4),\\
E_{\text{near }E}
&=E+\frac{\varepsilon^2V_0^2}{E-E_1}
+O(\varepsilon^4),\\
E_{\rm dark}&=E.
\end{aligned}
$$
These are exactly the three perturbative energies from part b through order $\varepsilon^2$.
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