How to cancel once-mentioned element in thermochemical reactions
We can make use of this formula: $$\Delta H = \Sigma ~\Delta H_f~(\text{products}) - \Sigma ~\Delta H_f~(\text{reactants}) $$

Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H_f$ for $\ce{O3}$. If you're interested, wikipedia has documented some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H_f = 0$.

Now that the $\Delta H_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H_f \ce{(NO2)} + 0) - (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3}))$$

So, from the equation above we can conclude that $$x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $$

Now write the master formula for the last reaction: $$(2 * \Delta H (\ce{NO}) + 0) - 2 * \Delta H_f (\ce{NO2})$$

What did we conclude from the last reaction? $x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H_f$ for $\ce{NO}$ &amp; $\ce{NO2}$?