\(\begin{array}{l}
1)\\
Ba{(OH)_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaOH\\
nBa{(OH)_2} = \dfrac{{40 \times 34,2\% }}{{171}} = 0,08\,mol\\
nN{a_2}S{O_4} = nBa{(OH)_2} = 0,08\,mol\\
\Rightarrow mN{a_2}S{O_4} = \dfrac{{0,08 \times 142}}{{14,2\% }} = 80g\\
2)\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
nF{e_2}{O_3} = \dfrac{{2,4}}{{160}} = 0,015\,mol\\
nFe{(OH)_3} = 2nF{e_2}{O_3} = 0,03\,mol\\
\Rightarrow x = mFe{(OH)_3} = 0,03 \times 107 = 3,21g
\end{array}\)