a,
$n_{CO_2}=\dfrac{5,824}{22,4}=0,26(mol)$
Đặt CTTQ 2 ankan là $C_nH_{2n+2}$
Bảo toàn $C$: $n.n_{\text{ankan}}=n_{CO_2}$
$\to n_{\text{ankan}}=\dfrac{0,26}{n}(mol)$
$\overline{M}=\dfrac{3,84n}{0,26}=14,77n=14n+2$
$\Leftrightarrow n=2,6$
Vậy CTPT 2 ankan là $C_2H_6$, $C_3H_8$
b,
Đặt $x$, $y$ là mol $C_2H_6$, $C_3H_8$
$\to 30x+44y=3,84$
Bảo toàn $C$: $2x+3y=0,26$
Giải hệ: $x=0,04; y=0,06$
$\%m_{C_2H_6}=\dfrac{0,04.30.100}{3,84}=31,25\%$
$\%m_{C_3H_8}=68,75\%$