I've attempted this problem but not sure if i'm on the right track.

Use the enthalpy diagram provided above and apply Hess’s Law to determine the standard enthalpy of formation for C12O36H20N12 (s) using the results from part (a) and the following values:
The standard enthalpy of formation of gaseous carbon dioxide is -393.5 kJ/mol
The standard enthalpy of formation of liquid water is -286 kJ/mol
(Strictly speaking, standard enthalpy is defined for a reaction that occurs at 25 °C, but for the purpose of this question you may ignore the difference in temperature.)

My attempt:
So i got that part a which asked for the heat evolved during the decomposition reaction of C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l) the molar enthalpy change is -756.5kJ.

I use delta h(rxn)=delta H(products-reactants)
The product is C12O36H20N12 (s) which is part a and is -756.5
the reactants are 12( -3935.5kJ)+10(-286kJ) with N2(g) and O2(g) being in their standard which is just zero.
Therefore we have -756.5kJ-(12( -3935.5kJ)+10(-286kJ)) which equals 6822.5kJ for the standard enthalpy of formation for C12O36H20N12. hints/tips are greatly appreciated

Your attempt at using Hess's Law to determine the standard enthalpy of formation for C12O36H20N12 (s) is on the right track. Here's a step-by-step guide on how to calculate the enthalpy of formation using Hess's Law:

Step 1: Write out the balanced equation
The balanced equation for the decomposition reaction is:
C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l)

Step 2: Determine the molar enthalpy change for this reaction
From part (a), you mentioned that the molar enthalpy change for this reaction is -756.5 kJ.

Step 3: Determine the molar enthalpy change for the formation reaction
The formation reaction can be written as:
C12O36H20N12 (s) → 12CO2 (g) + 6N2 (g) + 10H2O (l)

Since the standard enthalpy of formation of carbon dioxide is given as -393.5 kJ/mol and the standard enthalpy of formation of water is given as -286 kJ/mol, we can write the enthalpy change for the formation reaction as:
ΔH = (12 × -393.5 kJ) + (6 × 0 kJ) + (10 × -286 kJ)

Step 4: Apply Hess's Law
According to Hess's Law, the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for the individual reactions involved. So, we can write:
ΔH = ΔH formation - ΔH decomposition

Substituting the values into the equation, we get:
-756.5 kJ = [(12 × -393.5 kJ) + (6 × 0 kJ) + (10 × -286 kJ)] - ΔH formation

Simplifying the equation, we find:
ΔH formation = [(12 × -393.5 kJ) + (6 × 0 kJ) + (10 × -286 kJ)] - (-756.5 kJ)

Step 5: Calculate the enthalpy of formation
Evaluate the equation to find the enthalpy of formation:
ΔH formation = -4684.5 kJ - (-756.5 kJ)

ΔH formation = -4684.5 kJ + 756.5 kJ = -3928 kJ

Therefore, the standard enthalpy of formation for C12O36H20N12 (s) is -3928 kJ/mol.

Remember, it's always a good idea to double-check your calculations and make sure you've followed all the steps correctly.