\(\begin{array}{l}
2)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
2{C_6}{H_5}OH + 2Na \to 2{C_6}{H_5}ONa + {H_2}\\
n{H_2} = \dfrac{{0,56}}{{22,4}} = 0,025\,mol\\
{C_2}{H_5}OH(a\,mol),{C_6}{H_5}OH(b\,mol)\\
a + b = 0,05\\
46a + 94b = 3,74\\
\Rightarrow a = 0,02;b = 0,03\\
\% m{C_2}{H_5}OH = \dfrac{{0,02 \times 46}}{{3,74}} \times 100\% = 24,6\% \\
\% m{C_6}{H_5}OH = 100 - 24,6 = 75,4\% \\
3)\\
{C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH + 2C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
nC{O_2} = nCaC{O_3} = \dfrac{{40}}{{100}} = 0,4\,mol\\
\Rightarrow n{C_6}{H_{12}}{O_6} = \dfrac{{0,4}}{2} = 0,2\,mol\\
H = 75\% \Rightarrow m{C_6}{H_{12}}{O_6} = \dfrac{{0,2 \times 180}}{{75\% }} = 48g
\end{array}\)