\(\begin{array}{l}
a)\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
nS{O_2} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol\\
= > n{H_2}S{O_4} = n{H_2}O = 2nS{O_2} = 0,08\,mol\\
BTKL\\
mhh + m{H_2}S{O_4} = m + mS{O_2} + m{H_2}O\\
m = 5,03g\\
b)\\
hh:Al(a\,mol),Zn(b\,mol)\\
27a + 65b = 1,19\\
1,5a + b = 0,04\\
= > a = 0,02;b = 0,01\\
\% mAl = \dfrac{{0,02 \times 27}}{{1,19}} \times 100\% = 45,38\% \\
\% mZn = 100 - 45,38 = 54,62\%
\end{array}\)