$\begin{array}{l} a)2Mg+O_2 \to 2MgO\\b)n_{MgO}=\dfrac{8}{24+16}=0,2mol\\n_{Mg}=n_{MgO}=0,2mol\\m_{Mg}=0,2.24=4,8g\\c)n_{Mg}=0,2mol\\n_{O_2}=\dfrac{2}{32}=0,0625mol\\n_{Mg}=\dfrac{0,2}{2}=0,1mol>n_{O_2}=0,0625mol\\n_{Mg(pu)}=2.0,0625=0,125mol\\n_{Mg(du)}=0,2-0,125=0,075mol\\m_{Mg(du)}=0,075.24=1,8g\\n_{MgO}=2.0,0625=0,125mol\\m_{MgO}=0,125.(24+16)=5g\end{array}$