In BrF5, order to obtain pentavalency, some of electrons are shifted to 4d-orbitals. Two of p-orbitals become unpaired. During hybridisation one 4s, three 4p and two 4d orbitals take part. Therefore, hybridisation is sp3d2.
Br→[Ar]3d104s24p5 Xe→[Kr]4d105s25p6 In formation of XeF4, two electrons from 5p excited to vacant 5d orbitals. There are four unpaired electrons, which include two in 5p and two in 5d orbitals. Hence, sp3d2 hybridisation.