In the absence of other information, we assume we are in the frame of reference of the initial particle (i.e. treat the π⁰ as initially at rest before decay).

I’ll assume 140.106 eV/c² means 140x10⁶ eV/c² = 140MeV/c²

_________________________

1) Rest-energy of a π⁰ = m₀c² = 140MeV/c² x c² = 140MeV (just needed to change units).

Each of the decay particles must have half the rest energy of the π⁰.

Energy of each decay particle = 140MeV/2 = 70MeV

_________________________

2) The momentum of a photon is p = E/c.

p = 70MeV / c = 70MeV/c (just needed to change units).

_________________________

3) Electron rest mass m₀= 0.511MeV/c²

E² = p²c² + (m₀c²)²

70² = p²c² + (0.511)²

The electron’s total energy (70MeV) is much bigger than its rest-energy (0.511MeV). So (m₀c²)² is negligible compared to E².

p²c² = 4899.7

. . . .= 4900 near enough

pc = 70MeV

p= 70MeV/c

So the momentum is (practically) identical to the photon’s momentum in part 2).