What fraction of one edge-centred octahedral void lies in one unit cell of $fcc$?

In a cubic close packed structure,the fractional contributions of an atom at the corner and at the face in the unit cell are,respectively:

In the face-centered unit cell,the lattice points are present at

$A$ compound is formed by cation $C$ and anion $A$. The anions form hexagonal close packed $(hcp)$ lattice and the cations occupy $75 \%$ of octahedral voids. The formula of the compound is

$Ag$ crystallises in $fcc$ lattice. What is the total number of tetrahedral voids present in $540 \ g$ of $Ag$ metal (in $N_A$)? $(N_A = \text{Avogadro number}; Ag \text{ atomic weight} = 108 \ g \ mol^{-1})$

In an $fcc$ arrangement of $A$ and $B$ atoms, $A$ atoms are at the corners of the unit cell and $B$ atoms are at the face centers. If one of the $A$ atoms is missing from one corner, what is the simplest formula of the compound?