Your intuition is correct $-$ an asymmetry between $\alpha$ and $\beta$ is only possible if the two charges are different.
If you want a quantitative relationship between the two angles, the correct (read: the only viable) approach is via Gauss's law for the electric flux. From the geometry of the field, which is symmetric about the inter-charge axis, it is relatively easy to see that if you take the surface of revolution generated by the field line about the axis, then you get a cylinder-with-two-conical-ends which goes from one charge to the other:
By definition, the electric field is tangential to this surface at every point, which means that no field lines leave it, and the electric flux is confined inside it. That means, therefore, that the electric flux that leaves charge 1 into the surface must equal the electric flux that arrives at charge 2.
Moreover, we know how to relate these electric fluxes to the angles: close to charge 1, we can ignore the effect of charge 2, and then we just have the flat integral, i.e., the electric flux is the product of the charge times the solid angle spanned by the cone at its apex,
$$
\Phi_1 = q_1 \: \Omega_1,
$$
where the solid angle can be calculated explicitly as
\begin{align}
\Omega_1
& = \int_0^\alpha \int_0^2\pi \sin(\theta)\mathrm d\phi \:\mathrm d\theta
\\ & = 2\pi \big[-\cos(\theta)\big]_0^\alpha
\\ & = 2\pi(1-\cos(\alpha)).
\end{align}
Assuming that $q_1>0>q_2$, we can set $\Phi_1+\Phi_2=0$ and therefore
\begin{align}
2\pi(1-\cos(\alpha))|q_1| & = 2\pi(1-\cos(\beta))|q_2|
\\ \implies
\frac{1-\cos(\alpha)}{1-\cos(\beta)} & = \frac{|q_2|}{|q_1|}
.
\end{align}
This relationship then allows you to find any of the relevant quantities in terms of the other three, which is the most that you can do here.
Does that violate the rule that electric field lines (atleast those of
the coulomb force) can only start/end at charges/infinity?
No. Although the electric fields of each of the two equal charges vectorially equal zero at the mid point between the charges, the electric field lines do not "end". The lines are diverted from that space but still wind up terminating at a charge. The following link shows examples.
https://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines
The diagram below from the link demonstrates that the lines end at the charges.
A follow-up question - In light of the above case, how can we prove
that a point of zero electric field (a null point) in an "arbitrary
electrostatic field configuration" is necessarily a point of unstable
equilibrium?
Not quite sure what you mean by "prove", but clearly the field strength, which is proportional to the density of the field lines, is zero only at the mid point. A charge at any position not exactly at the mid point will experience a force and accelerate. Perhaps in that sense the situation of a charge at the mid point can be considered unstable.
Hope this helps.
Best Answer
Contrary to the comments, this problem can be solved analytically without the use of differential equations. Since this is a "homework-like" problem, but one requiring an unusual technique, I will describe the method in a bit more detail than I would otherwise. I will still leave the details of the calculations to you.
Let $x = 0$ denote the midpoint of the charges. Consider a closed surface consisting of three parts:
There is no charge enclosed by this surface, and by definition there is no flux through portion #3 of the surface since it follows the field lines. So by Gauss's Law we must have $$ \iint_1 \mathbf{E} \cdot d \mathbf{a} + \iint_2 \mathbf{E} \cdot d \mathbf{a} = 0. $$ In the limit $\epsilon \to 0$, the first integral will be dominated by the inward flux from the $+q$ charge, with the flux due to the $-q$ charge becoming negligible; so this flux can be calculated easily. The outward flux through surface #2 can also be calculated in terms of the unknown height $h$. Here we do have to take the effects of both charges into account; however, it is not hard to see that the fluxes through surface #2 due to the $+q$ charge and the $-q$ charge are equal, so that simplifies matters somewhat.
Take it from there.