You have to take into account the energy initially stored as the interaction energy of the particles. As mass and energy are just related by a factor of $c^2$, the initial rest mass of the "composite" particle is $m_1 + m_2 + E_{interaction}/c^2$. Indeed, note that in your calculations, the relativistic mass (or total energy) is not conserved (this can easily be seen in the reference frame two setup).
The expressions are not true in general. The first one should be $E^2 = m^2c^4 + p^2c^2$, and the momentum is in general $p = \gamma m v$. The rest energy is $E_0 = m c^2$ and it doesn't depend on the frame (by definition), and the kinetic energy is always $T=E-mc^2 = (\gamma - 1) mc^2$.
You are (understandably) confused because the question is not telling you that momentum is $mc$. You are being told that in a specific situation and in a specific frame, it just so happens that the momentum is equal to $mc$. You should be able to find the velocity from this, and then the kinetic energy.
Alright, since you're having trouble let's get our equations straight. First we define $\gamma$, which is a function of velocity $v$, as $1/\sqrt{1-v^2/c^2}$. The momentum $p$ of a particle with mass $m$ moving with velocity $v$ is given by $p = mv/\sqrt{1-v^2/c^2} = \gamma m v$. The expression $\gamma mv$ looks simpler but don't forget that $v$ is hidden inside $\gamma$.
There are two expressions for the energy. Obviously both are true and can be proved to be equal to each other; the only difference is whether you want the energy in terms of $p$ or $v$. So we have $E^2 = p^2c^2 + m^2c^4$ and $E = mc^2/\sqrt{1-v^2/c^2} = \gamma m c^2$. Kinetic energy $T$ is defined as $T = E-mc^2 = (\gamma-1)mc^2$. As always, this depends on $v$ through $\gamma$.
All these equations are true in any frame. The quantities themselves (such as $v$, $p$ or $E$) change when you change frames, but they change in such a way that all the equations remain correct.
Now, you have been told that in some particular frame, a particle is moving with $p=mc$. This will not be true in general, since the formula for $p$ is $mv/\sqrt{1-v^2/c^2}$; it just so happens that in our situation, $v$ is such that $p = mc$. This is an equation you can use to find $v$; knowing $v$, you can use the formula for kinetic energy $T$ (which, don't forget, depends on $v$) and find what you are being asked for.
Best Answer
For Minkowski or Schwartzschild spacetimes, the quantity $$m\left(X^i\frac{dX^j}{d\tau} - X^j\frac{dX^i}{d\tau}\right)$$ is conserved for masses following geodesic trajectories. It results from the existence of some Killing vectors.
In the Minkowski spacetime, the geodesics are straight lines, and it is the trivial fact that the relativistic angular momentum is just the distance to the line multiplied by the linear relativistic momentum (that is also conserved).
In the Schwartzschild spacetime, it means that the conservation of angular momentum of classical eliptical orbits is an approximation to the conservation of the relativistic angular momentum. Here it is supposed one big mass M, and only one small orbiting mass m, where M>>m.