 Solution

All three processes are weak charged current transformations between neighboring nuclear isobars. Neglecting neutrino masses, and writing $M_a(A,Z)$ for a neutral atomic mass, their Q values are:

These formulae use the paper's natural units. With energies in MeV, beta-plus decay requires an atomic mass-energy difference exceeding $2m_e{c}^2=1.022\mathrm{MeV}$. The extra electron-mass subtraction arises from using neutral atomic rather than bare nuclear masses. Electron capture can occur when the atomic mass-energy difference is positive but below this threshold. For a particular capture shell or excited daughter, the available energy must also cover the electron-hole and nuclear excitation energies, and the small nuclear recoil. Small atomic binding corrections are suppressed in the displayed threshold formulae.

The question's fixed-$A$ family consists of nuclear isobars, rather than isotopes: isotopes have fixed $Z$. The semi-empirical mass formula gives a convex nuclear mass parabola as a function of $Z$ at fixed $A$. For odd $A$, nuclear pairing does not split the family into two branches, so there is typically one beta-stable isobar, at the discrete minimum. Neighboring members can lower their mass by changing $Z$ toward it.

For even $A$, nuclear pairing lowers even-even nuclei and raises odd-odd nuclei, producing two offset nuclear mass parabolas. There are often two beta-stable even-even isobars, separated by an odd-odd member to which neither can energetically decay. The number is not an exact universal rule: some even-$A$ chains have one or more than two such minima against single beta decay, and a few light odd-odd nuclei are stable. Stability against single beta decay also does not guarantee stability against double beta decay.