\(\begin{array}{l}
a)\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
nS{O_2} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Mg(a\,mol),Cu(b\,mol)\\
24a + 64b = 15,2\\
a + b = 0,3\\
 \Rightarrow a = 0,1;b = 0,2\\
\% mMg = \dfrac{{0,1 \times 24}}{{15,2}} \times 100\%  = 15,79\% \\
\% mCu = 100 - 15,79 = 84,21\% \\
b)\\
n{H_2}S{O_4} = 2nS{O_2} = 0,6\,mol\\
CM{H_2}S{O_4} = \dfrac{{0,6}}{{0,8}} = 0,75M\\
CMMgS{O_4} = \dfrac{{0,1}}{{0,8}} = 0,125M\\
CMCuS{O_4} = \dfrac{{0,2}}{{0,8}} = 0,25M\\
c)\\
nCa{(OH)_2} = 1,25 \times 0,2 = 0,25\,mol\\
T = \dfrac{{0,25}}{{0,3}} = 0,833\\
Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O(1)\\
CaS{O_3} + S{O_2} + {H_2}O \to Ca{(HS{O_3})_2}(2)\\
nS{O_2}(1) = nCaS{O_3}(1) = 0,25\,mol\\
nS{O_2}(2) = nCaS{O_3}(2) = 0,3 - 0,25 = 0,05\,mol\\
CMCaS{O_3} = \dfrac{{0,2}}{{0,8}} = 0,25M\\
CMCa{(HS{O_3})_2} = \dfrac{{0,05}}{{0,8}} = 0,0625M
\end{array}\)