$a)$ Gọi: $\left \{ {{n_{CuO}=x(mol)} \atop {n_{Al_{2}O_{3}}=y(mol)}} \right.$
$m_{hh}=m_{CuO}+m_{Al_{2}O_{3}}=22,2(g)$
⇔$80x+102y=22,2(1)$
$n_{HCl}=\frac{300.10,95}{100.36,5}=0,9(mol)$
$PTHH:CuO+2HCl→CuCl_{2}+H_{2}O$
$mol:$ $x$ →$2x$→$x$
$PTHH:Al_{2}O_{3}+6HCl→2AlCl_{3}+3H_{2}O$
$mol:$ $y$ →$6y$ →$2y$
Có: $n_{HCl}=2x+6y=0,9(mol)(2)$
Giải hệ $(1)$,$(2)$ ⇒$\left \{ {{x=0,15} \atop {y=0,1}} \right.$
$m_{CuO}=\frac{0,15.80}{22,2}.100$%$=54,05$%
$m_{Al_{2}O_{3}}=100$%$-54,05$%$=45,95$%
$b)$ $m_{ddSPU}=22,2+300=322,2(g)$
$C$%$_{CuCl_{2}}=\frac{0,15.135}{322,2}.100$%$=6,28$%
$C$%$_{AlCl_{2}}=\frac{0,2.133,5}{322,2}.100$%$=8,29$%