a,

CT chung của 2 ancol: $C_nH_{2n+2}O$

$C_nH_{2n+2O}+\dfrac{3n}2O_2\xrightarrow{t^o} nCO_2+(n+1)H_2O$

$n_{O_2}=\dfrac{5,376}{22,4}=0,24\ (mol)$

$n_{H_2O}=\dfrac{4,68}{18}=0,26\ (mol)$

$\to \dfrac{3n}{2}.0,26=(n+1).0,24$

$\to n=1,6$

Vậy CTPT 2 ancol là $CH_4O:x\ (mol)$ và $C_2H_6O:y\ (mol)$

$CH_4O+\dfrac32O_2\xrightarrow{t^o} CO_2+2H_2O$

$C_2H_6O+3O_2\xrightarrow{t^o} 2CO_2+3H_2O$ 

$\to \begin{cases}1,5x+3y=0,24\\2x+3y=0,26\end{cases}\to \begin{cases}x=0,04\\y=0,06\end{cases}$

$\to \%m_{CH_4O}=\dfrac{0,04.32}{0,04.32+0,06.46}.100\%=31,68\%$

$\%m_{C_2H_6O}=100\%-31,68\%=68,32\%$

b,

$X\begin{cases}CH_3OH:0,04\\C_2H_5OH:0,06\end{cases}\xrightarrow{\ CuO\ } \begin{cases}HCHO:0,04\\CH_3CHO:0,06\end{cases}$

$HCHO\to 4Ag$

$CH_3CHO\to 2Ag$

$\to n_{Ag}=4.0,04+2.0,06=0,28\ (mol)$

$\to m_{Ag}=0,28.108=30,24\ (g)$