\(\begin{array}{l}
a)\\
{K_2}O + {H_2}O \to 2KOH\\
hh:{K_2}O(a\,mol),MgO(b\,mol)\\
m{\rm{dd}} = 94a + 190,6\\
nKOH = 2n{K_2}O = 2a\,mol\\
\Rightarrow \dfrac{{2a \times 56}}{{94a + 190,6}} = 5,6\% \Rightarrow a = 0,1\,mol\\
\Rightarrow mMgO = 20 - 0,1 \times 94 = 10,6g\\
\% mMgO = \dfrac{{10,6}}{{20}} \times 100\% = 53\% \\
b)\\
m{\rm{dd}} = 170 + 30 = 200g\\
C\% NaCl = \dfrac{{30}}{{200}} \times 100\% = 15\% \\
V{\rm{dd}} = \dfrac{{200}}{{1,2}} = \dfrac{{500}}{3}(ml) = \dfrac{1}{6}(l)\\
nNaCl = \dfrac{{30}}{{58,5}} = \dfrac{{20}}{{39}}\,mol\\
{C_M}NaCl = \dfrac{{\dfrac{{20}}{{39}}}}{{\dfrac{1}{6}}} = 3M
\end{array}\)