\(\begin{array}{l}
2)\\
mNaOH = m(g)\\
m{\rm{dd}} = 500 + m(g)\\
\dfrac{m}{{500 + m}} = 25\% \\
= > m = \dfrac{{500}}{3}(g)\\
m{\rm{dd}} = 500 + \dfrac{{500}}{3} = 666,67g\\
3)\\
n{H_2}S{O_4} = \dfrac{{98}}{{98}} = 1\,mol\\
V{H_2}S{O_4} = \dfrac{1}{{0,2}} = 5l
\end{array}\)