$n_{Cu}=6,4/64=0,1mol$
$2Cu+O_2\overset{t^o}{\longrightarrow}2CuO$
$a/$
Theo pt :
$n_{O_2}=1/2.n_{Cu}=1/2.0,1=0,05mol$
$⇒V_{O_2}=0,05.22,4=1,12l$
$b/$
$2KClO_3\overset{t^o}{\longrightarrow}2KCl+3O_2$
Theo pt :
$n_{KClO_3}=2/3.n_{O_2}=2/3.0,05=\dfrac{1}{30}mol$
$⇒m_{KClO_3}=\dfrac{1}{3}.122,5≈4,083g$