$n_{NaOH}=2n_{Na_2CO_3}=0,28(mol)$

$\to n_F=n_{OH}=0,28(mol)$ (vì không có este phenol)

$\to n_{H_2}=\dfrac{n_{OH}}{2}=0,14(mol)$

$\to m_F=10,36+0,14.2=10,64g$

$\overline{M_F}=38$

$\to F$ gồm $CH_3OH$ ($0,16$ mol), $C_2H_5OH$ ($0,12$ mol)

$m_{\text{muối}}=20,56+0,28.40-10,64=21,12g$

Muối gồm $HCOONa$ ($a$ mol), $C_2H_3COONa$ ($b$ mol), $CH_2$ ($c$ mol)

$HCOONa+0,5O_2\to \dfrac{1}{2}CO_2+\dfrac{1}{2}H_2O+\dfrac{1}{2}Na_2CO_3$

$C_2H_3COONa+3O_2\to \dfrac{5}{2}CO_2+\dfrac{1}{2}Na_2CO_3+\dfrac{3}{2}H_2O$

$CH_2+1,5O_2\to CO_2+H_2O$

$\to \begin{cases} a+b=0,28\\ 0,5a+3b+1,5c=0,34\\ 68a+94b+14c=21,12\end{cases}$

$\to \begin{cases} a=0,2\\ b=0,08\\ c=0\end{cases}$

Vậy hh este có: $\begin{cases} HCOOCH_3: 0,16\\ HCOOC_2H_5: 0,04\\ C_2H_3COOC_2H_5: 0,08\end{cases}$

Vậy $\%m_{HCOOCH_3}=46,69\%$