The protons that are different in the molecules will give you knowledge of their signals and where they should appear in the spectrum:
eugenol has an alcohol O-H which appears as a broad singlet in the spectrum
acetyleugenol has no alcohol O-H, but has an acetyl CH3 which will appear as a singlet.
Neither wll be coupled.
In the CO2, one of your structures is drawn incorrectly (the structure on the right, two Ha protons are bound to a carbon with three other bonds).
In all structures, you've drawn the Ha protons, but not the proton that undergoes 3J coupling with Hc and Hd (the hydrogens at the terminal end of the alkene). It is this missing proton, that splits Hc and Hd (to give overlapping doublet of doublets) and is split to give the complex pattern that you've labelled Hf.
I annotated your spectrum (CO2) and fixed the incorrect drawings.
The 5.4ppm peak is from the eugenol O-H. The 2.8ppm singlet is from the -CH3 on the acetyl functional group in acetyleugenol.
The hydrogens have already been integrated. A computer program in the NMR machine measures the integrals of the curves (the area under each curve) and prints this number at the bottom of the spectrum. The numbers are never exact, you have to assign integration comparatively within a spectrum:
once you find a peak representing a known number of hydrogens, you can use its integration as a basis to derive the integration of other peaks.
I don't understand this method of calculating the mole ratio of the hydrogens? The height of the integral does not offer you any useful information.
Are you sure this is the method you were given?