The question requires us to compare two neutralization reactions between NaOH and HCl, given the concentrantio of HCl and the amount of NaOH used, and explain why a smaller amount of NaOH was used for one of the reactions.
From the table, we can gather the following information regarding reactions 1 and 3:
Reaction 1: 20 drops of 0.1 M HCl + 42 drops of NaOH
Reaction 3: 20 drops of 0.05 M HCl + 43 drops of NaOH
Just comparing the concentrantion of HCl used in each reaction, we can see that for Reaction 3, a less concentrated HCl solution was used, although the volume of solution was the same in both cases.
Remember that the amount of a compund, in moles, can be calculated multiplying the solution concentration (in mol/L, for example) by its volume (in L), therefore if we are using the same volume but different concentrations, there will be a different amount of HCl for reaction 1 and 3 (smaller for 3 because the concentration is smaller).
Since we are talking about a neutralization reaction and the same 0.1 M NaOH solution was used for both reaction 1 and reaction 3, we should expect that a smaller amount of NaOH should be used for reaction 3 because the effective amount of HCl in the reaction was smaller. Therefore, we could answer the question "why the smaller amount of NaOH was required to neutralize the acid in Reaction 3" by comparing the concentration of HCl solutions - a smaller concentration in reaction 3 leads to a smaller amount of NaOH necessary to neutralize it.
On the other hand, the observations presented in table 2 say otherwise: more volume of NaOH was counted for reaction 3 compared to reaction 1.
To explain this, we could consider that an experimental error occured (for example, the wrong number of drops was counted for reaction 1, or the final point of the reaction - when it turns "pinkish" - was not the same for all reactions).