After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.
Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.
Part (c)
Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.
It can be written as
$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$
where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.
$\frac{k_2}{\omega}$ simply shifts the center of oscillation.
For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore
$$E^2 = (k_3-eA)^2 - \omega + (2 n_1 + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$
This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.
Part (e)
For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.
$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$
This leads to
$$E^2 = (k_3-eA)^2 + \omega + (2 n_1' + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$
Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.
These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.
We have however missed a special case where $n_1=0$. In this case, $E = \pm(k_3 - eA)$, and (19.7a) and (19.7b) reduce to
$$ \begin{equation} \begin{split} \phi_1' &= (k_2 - \omega x^1) \phi_1 \\ \phi_2' &= i(E-k_3+eA)\phi_1- (k_2 - \omega x^1) \phi_2. \end{split} \end{equation} $$
Now $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes. We have
Substituting this into (19.7a), where we can set $k_2$ to zero, we have
$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$
So we reach the same conclusion, but through a very different argument.