Thermochemistry Problem - How did they get that volume?
I think most of your issues regarding this question will be solved if you actually put the states in your chemical equations, so it becomes: $$\ce{CH4(g) + 2O2(g) -&gt; CO2(g) + 2H2O(l)}$$ It is important to note that all the substances are in the gaseous state except for water which is a liquid. Therefore your application of Gay-Laussac's law is incorrect here as the number of moles of gaseous reactants is not equal to number of moles of gaseous products. Thus the initial and final volume shouldn't be the same. This is also why in the question they have stated to assume that the volume of water is negligible as water is a liquid which has very little volume compared to a gas.

The change in volume can be calculated by noting that 3 moles of gas reacts to form only 1 mole of gas. Since there is 2L of methane, that means that the volume is going to decrease by 4L. 

Therefore: $\mathrm{V_f = V_i - 2.0L}$