\(\begin{array}{l}
a)\\
m_{(C_6H_{10}O_5)_n}=100.81\%=81(g)\\
\to n_{(C_6H_{10}O_5)_n}=\frac{81}{162n}=\frac{1}{2n}(mol)\\
(C_6H_{10}O_5)_n\to nC_6H_{12}O_6\to 2nC_2H_5OH\\
\to n_{C_2H_5OH}=2n.n_{(C_6H_{10}O_5)_n}=2n.\frac{1}{2n}=1(mol)\\
\to n_{C_2H_5OH(tt)}=1.75\%=0,75(mol)\\
\to V_{C_2H_5OH\,nc}=\frac{0,75.46}{0,8}=43,125(ml)\\
b)\\
V_{dd\,ruou}=\frac{43,125.100}{45}\approx 95,83(ml)\\
c)\\
V_{C_2H_5OH\,nc}=\frac{20.45}{100}=9(ml)\\
V_{H_2O}=20-9=11(ml)\\
n_{C_2H_5OH}=\frac{9.0,8}{46}=\frac{18}{115}(mol)\\
n_{H_2O}=\frac{11}{18}(mol)\\
2Na+2C_2H_5OH\to 2C_2H_5ONa+H_2\\
2Na+2H_2O\to 2NaOH+H_2\\
\to \sum n_{H_2}=\frac{1}{2}\left(n_{H_2O}+n_{C_2H_5OH}\right)=\frac{1589}{1035}(mol)\\
\to V_{H_2}=\frac{1589}{1035}.22,4\approx 34,4(l)
\end{array}\)