Use this for part (a)

N2H4(g) +2 H2O2(l) -> N2(g) + 4 H2O(g)

a) Calculate the enthalpy change, Delta{H}^{o}, for the above reaction using the enthalpy of formation values from the following table.

Delta{H}^{o}f (kJ/mol)

N2H4(g)                                                 95.40

H2O2(l)                                                  -187.8

H2O(g)                                                   -241.8

Hi
This question is very simple
Look at the balanced equation and sum up all enthalpies of formation of the reactants [(95.40+(2*(-187.)] .Now subtract this value from the sum of enthalpies of the  products [0+4*(-241.]
Take care of the units and remember for  elemental state  like N2 enthalpy of formation  is zero.
final answer will be : [0+4*(-241.]-[(95.40+(2*(-187.)] kJ/mol

That makes sense! Thank you!

Another question which I do not understand is,
Parts (b) through (c) pertain to the combustion of propane according to the process outlined below.

C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)   Delta{H}^{o}= -2043.9 kJ/mol

b) Calculate the amount of heat that is released when 6.78 g of propane, C3H8, is burned under standard conditions.


C) Explain why the combustion of C3H8 in the presence of O2 is exothermic in terms of bond energies.

For part b
Balanced equation is given
so you need to convert
6.78 g of propane----> moles of propane ----> enthalpy
so divide 6.78 g with the molar mass of propane and multiply it with the enthalpy value given for a mole in the reaction.

Exothermic means heat is released .It means during combustion of propane energy released during formation of  bonds of the products is more than energy required to break the bonds of the reactant. This you can prove also by subtracting the sum of the bond enthalpies of the products from the sum of the bond enthalpies of the reactants.

Thank you!