What happens in string theory beyond the Hagedorn temperature? What happens in string theory when the temperature exceeds the Hagedorn temperature? Is that even possible? If yes, what is the nature of the phase transition and the phase beyond that? What happens to spacetime in the new phase?
 A: The density of states for a string with respect to modes n is
$$
\eta(n)~\sim~ exp(4\pi n \sqrt{\alpha’})
$$
that defines a partition function
$$
 Z~\simeq~\int\eta(n)exp(-n/T)dn.  
$$
The temperature is computed by $1/T~=~\partial Z/\partial n$ and the path integral diverges for a temperature greater than 
$$
T_H~=~4\pi\sqrt{\alpha’}
$$
which is the Hagedorn temperature.  This is proportional to the reciprocal of the string length. The entropy of the system is the logarithm of the density of states the $S~\sim~ 1/nT_H$, which in the large $n$ limit is zero.  The modes number is given by $n~=~1/(\sqrt{d}M_s)$, for $d$ the number of degrees of freedom and $M_s$ the string mass
At this point there is new physics.  Hagedorn temperature occurs when physics scales below the string length, and pertains to a domain between the string length and the Planck length.  This the UV limit of string theory, which has correspondences with the IR limit by holography.  A gas of strings percolate into a single system (one large string), which has a correspondence with the Lorentz contraction and “spreading”of strings on the event horizon of a black hole at the IR limit.  At or near the Hagedorn temperature there is likely to be quantum fluctuations which exchange the various string types.  This in M-theory corresponds to the wrapping-mode T-duality, modular S-duality and such structures which exchange string types.  
I would conjecture that just as with the meson string, two quarks connected by a chromodynamic flux tube, strings are similar, and above the Hagedorn temperature become “gas” of D0-branes or particles.  The endpoints of strings (Chan-Paton factors or Dirichlet conditions on Dp-branes) become a plasma of D0-partons.
A: This depends a bit what you mean by string theory. I will take this to mean perturbative string theories. Also, my opinion below does not reflect a complete consensus, but I think it is the majority view. There is also lots of work on this, quite a bit of it before my time, I might be missing something.
So, in free string theories, there is exponential density of states at high energies. This means there is a “Hagedorn” temperature, a temperature beyond which the partition function $Tr(e^-{\beta H})$ does not exist.  The same appears to be happening in QCD, where there is an exponential density of almost stable resonances in some limit (the large N limit) - this is the context in which Hagedorn was discussing this temperature. 
It is an interesting question what happens as you heat up the system and approach this temperature. One has to distinguish two cases: the string coupling is precisely zero, or the string coupling is small but finite (which of course is the more sensible case).
At zero coupling, as you heat the system, you have competition between two situations. One is the conventional picture: gas of strings with different energy levels which are thermally distributed. On the other hand you can have a single excited string, which as a single object does not have a thermodynamic description. The reason you can have competition between such situations is because the single string has many “internal” degrees of freedom, so it can individually have large density of states. What happens near the Hagedorn temperature is that as you put more energy into the system, you excite a single string rather than change the thermal distribution of many strings. You can see the breakdown of the thermodynamic description because the specific heat of the system, which is a measure of the strength of the thermal fluctuations, diverges near the Hagedorn temperature.
If the coupling is finite, no matter how small, we have to consider interactions between the strings. The results of the classic paper by Atick and Witten suggest that well before you reach the Hagedorn temperature, the system will go through a first order phase transition and never really reach the Hagedorn temperature. The nature of the new phase was not clarified in that paper, and it probably depends on which string theory it is precisely. One universal expectation is that since strings attract gravitationally, the phase transition involves gravitational collapse of the system.
