a,
$n_{CO_2}=\dfrac{33}{44}=0,75(mol)$
$n_{H_2O}=\dfrac{13,5}{18}=0,75(mol)$
$n_{CO_2}=n_{H_2O}\to A$ có CTTQ $C_nH_{2n}$
$n=\dfrac{n_{CO_2}}{n_A}=3$
Vậy CTPT $A$ là $C_3H_6$
b,
$A$ mạch hở nên CTCT là: $CH_2=CHCH_3$
$CH_2=CHCH_3+H_2\xrightarrow{{Ni, t^o}} CH_3-CH_2-CH_3$ (propan)
$CH_2=CHCH_3+Br_2\to CH_2Br-CHBr-CH_3$ (1,2-đibrompropan)
$CH_2=CHCH_3+HBr\to CH_3-CHBr-CH_3$ (2-brompropan)
$n CH_2=CHCH_3\xrightarrow{{t^o, p, xt}} (-CH_2-CH(CH_3)-)_n$ (PP)