\(\begin{array}{l}
1)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
n{H_2} = \frac{{5,6}}{{22,4}} = 0,25mol\\
hh:Mg(a\,mol)Al(b\,mol)\\
24a + 27b = 5,1\\
a + 1,5b = 0,25\\
= > a = b = 0,1\,mol\\
\% mMg = \frac{{2,4}}{{5,1}} \times 100\% = 47,06\% \\
\% mAl = 100 - 47,06 = 52,94\% \\
b)\\
m = mAlC{l_3} + mMgC{l_2} = 0,1 \times 133,5 + 0,1 \times 95 = 22,85g\\
c)\\
m{\rm{dd}}HCl = \frac{{0,5 \times 36,5}}{{3,65\% }} = 500g\\
m{\rm{dd}}spu = 5,1 + 500 - 0,25 \times 2 = 504,6g\\
= > C\% \\
2)\\
C{l_2} + 2NaI \to 2NaCl + {I_2}\\
nC{l_2} = 0,2\,mol\\
nNaI = 0,5\,mol\\
= > nNaI\, = 0,5 - 0,2 \times 2 = 0,1\,mol\\
nNaCl = 0,2 \times 2 = 0,4\,mol\\
m = 0,4 \times 58,5 + 0,1 \times 150 = 38,4g
\end{array}\)