With the information given, we have to assume the calorimeter is "perfect", and no thermal energy is lost to either the calorimeter itself or to the surroundings in order to answer the questions.
The specific heat capacity of a material tells you how much energy it takes to raise the temperature of a unit quantity of the material by 1 degree. For instance, you are told that the specific heat capacity of water is 4186 J /kg∙°C. All that means is that it takes 4186 joules of energy to increase the temperature of 1 kg of water by 1°C.
In the first problem you have 1000ml (= 1 kg) of water at 48°C. A hot object is added and the water's temperature increases to 56°C at equilibrium, for a change in temperature of ΔT = +8°C. The energy required to do this is then:
q = m*C*ΔT, where m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature. In this case:
q = (1 kg)*(4186 J /kg∙°C)*(8°C) = 3.349*10^5 J. Given that the we are only given input values to 2 significant digits (at best, I'll assume there's exactly 1 kg of water in the calorimeter), we should round this 3.3*10^5 J
The second question is quite similar, and but in this case you are given q, the energy added to the water, and you have to calculate ΔT. Using the same formula:
810 J = (0.5 kg)*(4186 J /kg∙°C)*ΔT
ΔT = (810 J)/((0.5 kg)*(4186 J /kg∙°C)) = 0.387 °C. In this case, one of our pieces of input data only has 1 significant digit (0.5 kg), so this should be rounded to ΔT = 0.4 °C.