= Solution
Insert the expansions into the <energy eigenvalue> equation $\hat H\Psi_i=E_i\Psi_i$. Through first order,
$$
\begin{aligned}
\hat H\Psi_i&=\hat H^{(0)}\Psi_i^{(0)}
+\lambda\big(\hat H^{(0)}\Psi_i^{(1)}+\hat H^{(1)}\Psi_i^{(0)}\big)+O(\lambda^2),\\
E_i\Psi_i&=E_i^{(0)}\Psi_i^{(0)}
+\lambda\big(E_i^{(0)}\Psi_i^{(1)}+E_i^{(1)}\Psi_i^{(0)}\big)+O(\lambda^2).
\end{aligned}
$$
The zeroth-order coefficients reproduce part (i). Equating the coefficients of $\lambda$ and moving them to one side gives
$$
\boxed{(\hat H^{(0)}-E_i^{(0)})\Psi_i^{(1)}
+(\hat H^{(1)}-E_i^{(1)})\Psi_i^{(0)}=0.}
$$
Terms involving $\hat H^{(2)}$ first enter at order $\lambda^2$ and do not affect this equation of <first-order nondegenerate perturbation theory>.
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