How many tetrahedral voids are occupied in diamond? (in percentage)

Assertion : In close packing of spheres,a tetrahedral void is surrounded by four spheres whereas an octahedral void is surrounded by six spheres.
Reason : $A$ tetrahedral void has a tetrahedral shape whereas an octahedral void has an octahedral shape.

In a solid,oxide ions are arranged in $HCP$. Cations $A$ occupy one-sixth of $THV$ and cations $B$ occupy one-third of the $OHV$. The empirical formula of the compound is

$A$ solid contains elements $A$ and $B$. Anions of $B$ form $ccp$ lattice. Cations of $A$ occupy $50 \%$ of octahedral voids and $50 \%$ of tetrahedral voids. What is the molecular formula of the solid?

Atoms of element $B$ form a $ccp$ lattice and atoms of element $A$ occupy $\frac{1}{2}$ of the tetrahedral voids. What is the formula of the compound formed by elements $A$ and $B$?

In hexagonal systems of crystals,a frequently encountered arrangement of atoms is described as a hexagonal prism. Here,the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. $A$ space-filling model of this structure,called hexagonal close-packed $(HCP)$,is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally,the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be $r$.
$1.$ The number of atoms on this $HCP$ unit cell is
$(A)$ $4$ $(B)$ $6$ $(C)$ $12$ $(D)$ $17$
$2.$ The volume of this $HCP$ unit cell is
$(A)$ $24 \sqrt{2} r^3$ $(B)$ $16 \sqrt{2} r^3$
$(C)$ $12 \sqrt{2} r^3$ $(D)$ $\frac{64 r^3}{3 \sqrt{3}}$
$3.$ The empty space in this $HCP$ unit cell is
$(A)$ $74 \%$ $(B)$ $47.6 \%$ $(C)$ $32 \%$ $(D)$ $26 \%$
Give the answer for questions $1, 2$ and $3.$