Chemistry Problem Energetics

A calorimeter was calibrated by burning 2.00 g of methanol whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50g of propan-2-ol raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.
So far this is what I have managed to do:
Delta H = q/moles
-715 = q/0.0635 , which is the moles of methanol
q = -44.6875kJ
since, q = mc delta T
-44687.5J = m x 4.18 x 32.8
m = -325.938 cm3 but i know that the solution cannot be minus so its just 325.938cm3
Propan-2-ol:
q=mc delta T
q = 325.938 x 36.4
q = -49.5921 kJ
delta H = q/ moles of fuel
delta H = -49.5921/ (1.5/60)
delta H = -1983.684 kJ mol-1
so enthalpy of combustion of propan-2-ol is = -1984 kJ mol-1
I am still unsure if my calculations for enthalpy of combustion of propan-2-ol is correct. Also I am unsure how to calculate the heat capacity of the calorimeter??
Thanks

A calorimeter was calibrated by burning 2.00 g of methanol whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50g of propan-2-ol raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

So far this is what I have managed to do:
Delta H = q/moles
-715 = q/0.0635 , which is the moles of methanol
q = -44.6875kJ

since, q = mc delta T
-44687.5J = m x 4.18 x 32.8
m = -325.938 cm3 but i know that the solution cannot be minus so its just 325.938cm3

Propan-2-ol:
q=mc delta T
q = 325.938 x 36.4
q = -49.5921 kJ

delta H = q/ moles of fuel
delta H = -49.5921/ (1.5/60)
delta H = -1983.684 kJ mol-1

so enthalpy of combustion of propan-2-ol is = -1984 kJ mol-1

I am still unsure if my calculations for enthalpy of combustion of propan-2-ol is correct. Also I am unsure how to calculate the heat capacity of the calorimeter??

Thanks

They've asked you to calculate the value of c - what you've done is assumed c = 4.18 J/g/K (i.e the specific heat capacity of water), which has missed the point of the first bit of the question.
In fairness, they haven't specified the mass, so what you actually end up calculating is mc, where m is the mass of the calorimeter and c is the heat capacity of the calorimeter.
I would suggest you write that the magnitude of the enthalpy change is given by q/moles, so 715 kJ/mol = (q kJ)/(0.0625 mol) => q = 44.6875 kJ
Therefore mc = (44.6875 kJ)/(32.8 K) = 1.362... kJ/K (or about 1362 J/K)
From this and the rest of the data, you should get the same answer as above. This is because later, you've also remultiplied by the specific heat capacity of water, which is mathematically equivalent to multiplying by the correct value of mc.

As above, the calculations for methanol and propan-2-ol shared by the OP are where the problem with finding the specific heat capacity of the calorimeter originate. If they had been told that water was being heated and were asked to find the mass of water, then yes, I'd agree that their calculations are strictly correct. However, they appear to have become used to using c = 4.18 J/g/K as this is the standard practice at A level and so this has led to them assuming a value for c rather than calculating the correct one.
I agree with point 3, as it is ambiguous as to whether the 0.0635 moles* they have stated was calculated from more precise relative atomic masses or whether it is a typo and should instead read as 0.0625 moles. An A level examiner likely would deduct marks for transcription errors without working.
With point 4, if they were to do that, they'd just end up getting c = 4.18 J/g/K, which most probably is not the correct answer. I very much doubt that the calorimeter would have the required mass of 328 g, given the typical choices of material used to make them.
In point 5, it seems now you are also assuming that water is being heated. The assumption that the volume being equal in magnitude to the mass comes from the density of water being approximately 1 g/cm^3, which works to a good approximation when water or an aqueous solution is heated, but most probably is not appropriate in this case.
Point 6 is just a repeat of point 4.
*Edit: it couldn't possibly be due to use of more precise relative atomic masses - it would be more likely put down to use of an incorrect molar mass of methanol. Gotta love having a completely scrambled brain lol.