$2CH_3COOH+Mg\to (CH_3COO)_2Mg+H_2↑$
$n_{(CH_3COO)_2Mg}=\dfrac{1,42}{142}=0,01mol$
a.Theo pt :
$n_{CH_3COOH}=2.n_{(CH_3COO)_2Mg}=2.0,01=0,02mol$
$⇒C_{M_{CH_3COOH}}=\dfrac{0,02}{0,2}=0,1M$
b.Theo pt :
$n_{H_2}=n_{(CH_3COO)_2Mg}=0,01mol$
$⇒V_{H_2}=0,01.22,4=0,224l$
$c.CH_3COOH+NaOH\to CH_3COONa+H_2O$
Theo pt :
$n_{NaOH}=n_{CH_3COOH}=0,02mol$
$V_{NaOH}=\dfrac{n}{C_M}=\dfrac{0,02}{0,5}=0,04l$