a,
$n_{CH_3COOH}=\frac{5,8}{60}= \frac{29}{300} mol$
$2CH_3COOH+ Zn \to (CH_3COO)_2Zn+ H_2$
$\Rightarrow n_{H_2}= \frac{29}{600} mol$
$V_{H_2}=22,4.\frac{29}{600}= 1,08l$
b,
$V_{C_2H_5OH}=15,5.45\%=6,975ml$
$\Rightarrow n_{C_2H_5OH}=\frac{6,975.0,9}{46}= 0,136 mol$
$CH_3COOH+C_2H_5OH\buildrel{{H_2SO_4, t^o}}\over\rightleftharpoons CH_3COOC_2H_5+ H_2O$
$\Rightarrow $ Dư $0,136-\frac{29}{300}= 0,039 mol$ etanol
$m_{C_2H_5OH dư}= 46.0,039= 1,794g$