= Solution
In the original eigenbasis, $Q'=\begin{pmatrix}-1&i\epsilon\\-i\epsilon&1\end{pmatrix}$. Insert the proposed first-order state for <eigenvalue> near minus one. The first component gives $c_1=0$; the second gives $-ia_1+a_2=-a_2$, hence $a_2=ia_1/2$. Similarly for the <eigenvalue> near one, $c_2=0$ and $b_2=ib_1/2$. Normalization gives $|a_1|=|b_1|=1$, with their phases freely selectable. Choosing the original-state phase conventions,
$$
\boxed{a_1=b_1=1,\qquad a_2=b_2=i/2,\qquad c_1=c_2=0.}
$$
The states are orthogonal through first order because the two cross terms cancel.
Exactly, $(Q')^2=(1+\epsilon^2)I$, so putting $R=\sqrt{1+\epsilon^2}$ gives
$$
\boxed{q'_1=-R,\qquad q'_2=R.}
$$
For an <eigenvalue> $q'$, the first row requires the component ratio $v_2/v_1=-i(1+q')/\epsilon$, so the coefficient in the printed parametrization is \b[$\boxed{B=1}$]. With $D=\sqrt{(1+R)^2+\epsilon^2}$, smooth normalized choices are
$$
\boxed{|1'\rangle=\frac1D\binom{1+R}{i\epsilon},\qquad
|2'\rangle=\frac1D\binom{i\epsilon}{1+R}.}
$$
They have unit norm and zero inner product, and direct multiplication gives the two <eigenvalue> equations. For $\epsilon\ne0$, the printed scalar factors can be chosen $A_1=(1+R)/D$ and $A_2=i\epsilon/D$; the second is allowed a complex phase. The displayed ratio form is singular at zero, but the normalized vectors have the correct continuous limits there.
Expanding $R=1+\epsilon^2/2+O(\epsilon^4)$ and $D=2+O(\epsilon^2)$ gives $|1'\rangle=|1\rangle+i\epsilon|2\rangle/2+O(\epsilon^2)$ and $|2'\rangle=|2\rangle+i\epsilon|1\rangle/2+O(\epsilon^2)$, while the <eigenvalues> have no first-order shift. This recovers all six perturbative coefficients with the same phase convention.
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