\(\begin{array}{l}
a)\\
m_{X_2O_5}=76,4-0,3.160=28,4(g)\\
\to M_{X_2O_5}=2M_X+16.5=\frac{28,4}{0,2}=142(g/mol)\\
\to M_X=31(P)\\
\to X:\,photpho\,(P)\\
b)\\
\%m_{Fe_2O_3}=\frac{0,3.160}{76,4}.100\%\approx 62,83\%\\
\%m_{P_2O_5}=100-62,83=37,17\%
\end{array}\)