First Question: (edited for the annihilation-creation diagram)

The answer follows from the on-shell condition.
E1+m = E3+E4, p1 = p3+p4
This means that E1 and p1 can only be on-shell if p3=p4=0

Second Question:

General advice: First try to simplify things with the given relations before
starting to take squares and things like that. Then use the on-shell
relations to prove the given restrictions.

I suppose the '*' is the cross product here. The cross product means that
p3 and p4 are in the same direction, thus (p4x,p4y,p4z) = a(p3x,p3y,p3z)

This means that either [itex]a[/itex] must be 0 and thus p4=0 or p3=0.
There is another solution and that is [itex]a=1[/itex]. In this trivial case p4=p3
and they can be non-zero......

To which of the two diagrams contributing to Bhabha scattering are you
referring to. The electron positron scattering or the annihilation/creation one?

It holds for the x product....:rofl: (for the annihilation/creation diagram)

To which of the two diagrams contributing to Bhabha scattering are you
referring to. The electron positron scattering or the annihilation/creation one?