a,
X có CTTQ $C_nH_{2n+2}O$
$M_X=37.2=74= 14n+18$
$\Leftrightarrow n=4$
Vậy ancol X là $C_4H_{10}O$
b,
CTCT:
(1) $CH_2OH-CH_2-CH_2-CH_3$: butan-1-ol
(2) $CH_3-CHOH-CH_2-CH_3$: butan-2-ol
(3) $CH_2OH-CH(CH_3)-CH_3$: 2-metylpropan-1-ol
(4) $(CH_3)_3-COH$: 2-metylpropan-2-ol