Regarding enthalpy of a reaction
You should be careful to distinguish between the magnitude of enthalpy (or some other change in energy) and the sign of the property:


the magnitude tells you how large the change was (yes, self-evidently)
the sign tells you in which direction the change occurred:

a) $\text{system}\rightarrow \text{surroundings}$, $\Delta H&lt;0$, or exothermic
b) $\text{surroundings}\rightarrow \text{system}$, $\Delta H&gt;0$, or endothermic  


This is the typical convention used (not the only one).

In the case of the condensation of a gas, the magnitude can vary depending on the strength of intermolecular interactions in the liquid compared to the gas. The sign is negative because condensation at a constant temperature requires dissipation of heat to the surroundings. 


  So which one is it, where am I going wrong ??


That the enthalpy of $\ce{H2O(l)}$ is less than the enthalpy of $\ce{H2O(g)}$ means in the scenario you present that 
$$\Delta H_2 &lt; \Delta H_1$$
not the other way around. You could also start from the fact that condensation is exothermic (or vaporization endothermic) so that $H_g-H_l=-\Delta_{cond}H =\Delta_{vap}H&gt;0$$^\dagger$, from which $$\begin{align}H_g&amp;&gt;H_l\\ H_g-H_{reagents}&amp;&gt;H_l-H_{reagents}\\ \Delta H_1&amp;&gt;\Delta H_2\end{align}$$



$\dagger$ (Writing "$H_g$" is a bit "dangerous" because it suggests there is an absolute enthalpy scale (rather than relative) but you can assume that here you are using a common reference state for water).