\(\begin{array}{l}
1)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
n{H_2} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol \Rightarrow nK = 0,2 \times 2 = 0,4\,mol\\
\Rightarrow mK = 0,4 \times 39 = 15,6g\\
mFe = 17,6 - 15,6 = 2g\\
2)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
{K_2}O + {H_2}O \to 2KOH\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow nK = 2 \times 0,15 = 0,3\,mol\\
\% mK = \dfrac{{0,3 \times 39}}{{21,3}} \times 100\% = 54,93\% \\
\% m{K_2}O = 100 - 54,93 = 45,07\%
\end{array}\)