How to calculate the temperature from heat of neutralisation?
First, you need to recognize what the limiting reagent is.  Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($\pu{25mL}$ of $\pu{1.79M}$ is fewer moles than $\pu{50 mL}$ of $\pu{1.6M}$), HCl is the limiting reagent.
Second, given the amount of limiting reagent, how much heat will be released (or absorbed).  $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$
Third, you need to approximate that the solution has the heat capacity of water, which is $\pu{4.18 kJ/K L}$.
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average.  So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$
So finally calculate the amount the temperature of $\pu{75 ml}$ of &quot;water&quot; will increase when $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$ of heat is added, and add this to $\pu{25.55 ^\circ C}$.