How to determine the specific heat of methanol from calorimeter data?
You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at $\pu{24.7 ^\circ C}$ before the methanol was added and then it rose $\pu{3.8 ^\circ C}$ along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is $\pu{19.3 J/^\circ C}$ then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $ &quot;Joules absorbed by the calorimeter&quot;.
Then we simply apply the heat lost equals heat gained concept:
$$\begin{multline}
\pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\
  = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C}) 
    \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C} 
    \cdot \pu{3.8 ^\circ C}
\end{multline}$$
After rearranging, we get $S = \pu{2.36 J/(g * ^\circ C)}$.