Chemsitry Enthalpy Change Question Help

The moles of both ethanoic acid and sodium hydroxide are $\frac{2*25}{1000}$ = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.
Since $\Delta H = \frac{Q}{n}$, this means that $Q = \Delta H n$. Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use $Q=mc \Delta T$. Rearranging for $\Delta T$ would give $\Delta T = \frac{Q}{mc}$. Let's make a checklist of the terms in the equation:

So, we have everything to calculate the temperature change. $\Delta T = \frac{2805}{50*4.18}$ = 13.4 K.

Would you not calculate the mass of water produced in the reaction? I specifically remember being told that the mass is also the mass of water and not the mass of any solids/solutions?

No you wouldn't. The water is simply being formed, not undergoing any reaction; it's the 25 cm³ of ethanoic acid and the 25 cm³ of sodium hydroxide which are reacting and causing an enthalpy change.
What you were very much likely told was that the density of a solution can be assumed to be the density of water. The question wouldn't tell you to assume that the densities of both solutions are equal to that of water for nothing...