Why during annihilation of an electron and positron 2 gamma rays are produced instead of 1? $1 \gamma \rightarrow 1 e^- + 1 e^+$       (pair production)
Then why
$1 e^- + 1 e^+ \rightarrow 2 \gamma$       (annihilation of matter)
instead of
$1 e^- + 1 e^+ -> 1 \gamma$ ?
 A: The gamma photon should move with $c$, in any reference frame. Thus, it has to have a nonzero inertia.
But in the center of mass of the $e^-$ and the $e^+$, they have zero summed inertia.
Inertia is conserved, thus the resulting gamma photon should have zero inertia. It is impossible.
Ext:
If their spins are into the same direction, even the 2-photon annihillation is not possible on a different conservation law, the spin conservation. In this case, 3 photons (or more) are produced. The electrons (and positrons) have 1/2 spin, thus they can annihillate in +1/2, -1/2 or in +1/2, +1/2 configuration. Thus, their summed spin is either 0 or 1. The spin of the photons is 1, also they can have opposite or equal sign, thus their total spin want to sum up to 0 or 2. Thus, in the case if the electron and the positron have equal spin signum, the spins couldn't be conserved in a 2-photon annihillation.
A: The  answers given above are incomplete, probably because the OP is misleading. Everyone discussed the case of pair annihilation (or creation, if you wish to consider time-reverse processes) in vacuo, in which case the answers provided are correct.
However, being in vacuo is a useful generalization, but often is also an oversimplification. For instance pair creation/annihilation can occur inside an electric field (say, close to a nucleus), or inside a magnetic field (again, close to a nucleus or even an atom with non-vanishing magnetic moment), in which case one-photon processes are fully allowed, and quadri-momentum conservation is made possible by the presence of the nucleus and/or atom.
These are classical processes, widely discussed in the literature. For the case close to a nucleus, you may read Bethe and Heitler, 1934, Proc. Roy. Soc. London, A146, 83, and for magnetic field Zaumen, 1976,  ApJ, 210,776. Or summaries in Lang, Astrophysical Formulae, 1998, vol.I, pag. 433, and Meszaros, 1992, High-Energy radiation from magnetized neutron stars, U.Chicago Press, pag. 211.
A: This has a simple answer: the process $e^++e^-\to\gamma$ cannot satisfy both momentum and energy conservation at the same time. To see, this let's choose a reference frame in which the total momentum of the system is zero (that is, the electron has the opposite momentum to the positron). This reference frame is always possible to be chosen by a simple Lorentz transformation.
Now, the photon produced must have zero total momentum. However, this simply isn't possible, since photons must always travel at the speed of light. More importantly, if the produced photon has frequency $\omega$, then the momentum must satisfy $|\textbf{k}|=\hbar\omega/c$. Since $\hbar\omega$ is the energy of the photon, $\hbar\omega\geq 2m_e$ by energy conservation, and so $|\textbf{k}|>0$.
The $e^++e^-\to\gamma+\gamma$ process is perfectly allowed since, if the outgoing photons have opposite momenta, the center-of-mass frame can still be perfectly satisfied.
I hope this helped!
