In an electron-positron annihilation, in what direction are the photons released?

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles.

So assuming the spins of the electron-positron pair are averaged, and that you don't care about the photon polarization, if you define $\theta$ to be the angle the outgoing photons make with the direction of the incident particles, and you work in the COM frame, you get

\begin{align*} \left| \frac{d\sigma}{d \Omega}\right|\propto \frac{E^2 + p^2 \text{cos}^2\theta + 2m_e^2}{m_e^2 + p^2 \text{cos}^2\theta} - \frac{2m_e^4}{\left( m_e^2 + p^2 \text{cos}^2\theta \right)^2} \end{align*} Where $E$ is the energy of the incoming electron, and $p$ is its momentum, and $c\equiv 1$.

For very energetic $e^+ e^-$ this becomes $$ \left| \frac{d\sigma}{d \Omega}\right|\propto 2\text{csc}^2 \theta -1 $$

fun fact: for slow $e^+ e^-$ their De Broglie size will be big as to effectively set a lower limit on the impact parameter, which will make the formation of the meta stable bound state positronium much more likely. After that, the decay will depend on the specifics of the bound state.