Explanation

(a)
$XeF_{2},\ XeF_{4}\ and\ XeF_{6}$ can be directly prepared
$Xe + F_{2} + \underset{673\ K}{\overset{Ni\ tube}{\rightarrow}}\ XeF_{2};$
$Xe + 2F_{2}\underset{6\ atm}{\overset{673\ K}{\rightarrow}}\ XeF_{4}$
$Xe + 3F_{2}\ \underset{50 - 60\ atm}{\overset{523 - 573\ K}{\rightarrow}}\ XeF_{6}$
$XeO_{3}$ is obtained by the hydrolysis of $XeF_{6}$
$XeF_{6} + 3H_{2}O \rightarrow XeO_{3} + 6HF$