Cobalt-60 (mass 59.933817 u) decays by beta-minus to an excited state of nickel-60 (ground-state…

Step 1. Cobalt-60. $59.933817\ \text{u} - 59.930786\ \text{u} = 0.003031$ u. So $Q = 0.003031\ \text{u} \times 931.494\ \text{MeV/u} = 2.823$ MeV.

Step 2. The beta particle. The gamma rays carry $1.173 + 1.332 = 2.505$ MeV. The beta particle, the antineutrino and the tiny nuclear recoil share the rest, $2.823 - 2.505 = 0.318$ MeV. The beta particle has its largest energy when the antineutrino carries off almost none.

Step 3. Nitrogen-13. $13.005739\ \text{u} - 13.003355\ \text{u} = 0.002384$ u, which is $2.221$ MeV. Then $Q = 2.221\ \text{MeV} - 1.022\ \text{MeV} = 1.199$ MeV.