The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF₄, respectively, are:

The number of valence electrons in Xe is 8.
The number of valence electrons in O is 6.
The number of valence electrons in F is 7.
The total number of valence electrons in XeOF₄ is 8 + 6 + 4(7) = 42.
The Lewis structure of XeOF₄ is shown below:

   O
  / \
 F-Xe-F
  \ / 
   F F

Xe is the central atom, and it is bonded to four F atoms and one O atom. The formal charge of each atom is:
Xe: 8 - 4 - 4 = 0
F: 7 - 1 = 6
O: 6 - 2 - 2 = 0

The steric number of Xe is 5 (4 bonds + 1 lone pair). Therefore, the hybridization of Xe is sp3d.
The number of lone pairs of electrons on Xe is 1.
Therefore, the type of hybridization and the number of lone pairs of electrons of Xe in XeOF₄ are sp3d and 1, respectively.