a, 

$\overline{M}=23,8.2=47,6=14n$

$\Leftrightarrow a=3,4$ ($C_3H_6, C_4H_8$) 

Vậy 2 ancol là $C_3H_7OH$ (tạo a mol anken), $C_4H_9OH$ (tạo b mol anken)

CTCT: 

- Ancol 3C:

$CH_3-CH_2-CH_2OH$

$CH_3-CHOH-CH_3$

- Ancol 4C:

$CH_2OH-CH_2-CH_2-CH_3$

$CH_3-CHOH-CH_2-CH_3$

$CH_2OH-CH(CH_3)-CH_3$

$CH_3-C(OH)(CH_3)-CH_3$

$\Rightarrow \frac{42a+56b}{a+b}=47,6$

$\Leftrightarrow 5,6a=8,4b$

$\Leftrightarrow \frac{a}{b}=\frac{3}{2}$ 

Giả sử $a=3; b=2$

$\%m_{C_3H_7OH}=\frac{3.42.100}{3.42+2.56}= 52,9\%$

$\%m_{C_4H_9OH}=47,1\%$ 

b, 

Gọi 3x, 2x là mol $C_3H_7OH$, $C_4H_9OH$.

$\Rightarrow 46.3x+60.2x=6,56$

$\Leftrightarrow x= 0,025$

$n_{CO_2}= 3n_{C_3H_7OH}+ 4n_{C_4H_9OH}= 3.3x+4.2x= 0,425 mol$

$\Rightarrow m_{CO_2}=18,7g$

$n_{H_2O}= 4n_{C_3H_7OH}+ 5n_{C_4H_9OH}= 4.3x+5.2x= 0,55 mol$

$\Rightarrow m_{H_2O}=0,55.18=9,9g$