= Solution
Let $\Delta=E_+-E_->0$. Hermiticity gives
$$
\langle-|\Delta H|+\rangle=-i\lambda,
$$
so in the ordered basis $(| -\rangle,|+\rangle)$,
$$
H_0+\Delta H=
\begin{pmatrix}
E_-&-i\lambda\\
i\lambda&E_+
\end{pmatrix}.
$$
For a <nondegenerate energy eigenvalue>[nondegenerate unperturbed level], the <first-order nondegenerate perturbation theory> formulas are
$$
E_n^{(1)}=\langle n|\Delta H|n\rangle,
\qquad
|n^{(1)}\rangle
=\sum_{m\ne n}|m\rangle
\frac{\langle m|\Delta H|n\rangle}{E_n-E_m}.
$$
The diagonal matrix elements vanish, so both linear energy corrections are zero. The state corrections are
$$
| -^{(1)}\rangle
=|+\rangle\frac{i\lambda}{E_--E_+}
=-\frac{i\lambda}{\Delta}|+\rangle,
$$
$$
|+^{(1)}\rangle
=|-\rangle\frac{-i\lambda}{E_+-E_-}
=-\frac{i\lambda}{\Delta}|-\rangle.
$$
Thus, through linear order,
$$
\boxed{
\begin{aligned}
E_-^{\rm pert}&=E_-+O(\lambda^2),&
|-\rangle_{\rm pert}&=|-\rangle-\frac{i\lambda}{\Delta}|+\rangle+O(\lambda^2),\\
E_+^{\rm pert}&=E_++O(\lambda^2),&
|+\rangle_{\rm pert}&=|+\rangle-\frac{i\lambda}{\Delta}|-\rangle+O(\lambda^2).
\end{aligned}}
$$
Normalization changes only at quadratic order.
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