In the fluorite structure of $\mathrm{CaF}{}_2$, the coordination numbers of $\mathrm{Ca}{}^{2+}$ and $\mathrm{F}{}^{-}$ are, respectively,

$\mathrm{Ca}{}^{2+}$ forms an fcc lattice and $\mathrm{F}{}^{-}$ fills all tetrahedral voids. Each $\mathrm{F}{}^{-}$ is surrounded by 4 $\mathrm{Ca}{}^{2+}$ (tetrahedral), and by the 1:2 ratio each $\mathrm{Ca}{}^{2+}$ is surrounded by 8 $\mathrm{F}{}^{-}$. The answer is 8 and 4.