Good, looks like you're calculating the composition correctly. Do you notice anything about the probability values - and Q - in the limit of high temperature? Here's a hint: Q starts at 6 (T = 0) and approaches 12 (T = infinity). PA, PB, and PC approach values of 0.500, 0.333, and 0.167, respectively. This is telling you something about the impact of temperature on the population of available states.
For internal energy, it looks like you're doing something wrong. For one thing, I know what the value is, but even without that: at T = 0, all of the quantum objects are a state with E = 0, so the total energy must be zero. At higher temperatures, some objects are, on average, in states with E > 0. So the energy of your system a T > 0 should have an average E > 0. E(T > 0) > E (T = 0). The energy should be the statistically weighted average of all particles in the system - so, for example, in a two state system that has half population of the ground state and half population of the excited state, the internal energy will be halfway between the energy of the upper state and the lower state. From this information alone, you should be able to tell what it will be for you specific system here.
Actually, you should be calculating your energy (call it U) value first, then your entropy value, which can conveniently be calculated from U.
In terms of Q, the average internal energy, U, for a body of N objects, can be expressed basically as
[tex]U = - \frac {N}{Q} \frac {dQ}{d\beta}[/tex]
(Warning, that's based off memory. You may want to consult your physical chemistry book to make sure.) Really, it's a partial derivative but effectively the same for your purposes here.
For entropy, there are a couple of different ways to express it. I can't remember that equation off the top of my head, or the value I got, but I can check for you tomorrow morning.