Answer
See answers.
Work Step by Step
a. By charge conservation, if 2 beta particles are emitted, each with a charge of -1, then in the parent nucleus, 2 neutrons must change into protons. The mass number stays the same, but the atomic number increases by 2. The daughter nucleus is $^{96}_{42}Mo$.
$$^{96}_{40}Zr\rightarrow\;^{96}_{42}Mo +2\beta^-$$
b. The conservation of lepton number would be violated by neutrinoless double beta decay, since the 2 $\beta$ particles (electrons) are leptons, each with lepton number of +1. Protons and neutrons have lepton number of 0.
c. If the parent nucleus $^{96}_{40}Zr $ were to do 2 “normal” beta decays, it would also emit 2 electron antineutrinos, each with lepton number of -1. It could decay to $^{96}_{42}Mo$ and conserve lepton number.
