To begin with, a disclaimer: The approach here only takes into account the equilibria that the question itself cares about, namely $\ce{CaF2}$ dissociation and $\ce{F-/HF}$ acid-base. As pointed out in DavePhD's answer as well as Linear Christmas's comment, for a realistic treatment of the system, extra parameters must be taken into account.
A more accurate calculation would involve setting up five equations for five unknowns and solving them. The first three equations come from the data you provided. All concentrations are in $\pu{mol dm-3}$.
$$\begin{align}
[\ce{Ca^2+}][\ce{F-}]^2 &= 3.9 \times 10^{-11} \tag{1} \\
\frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]} &= 6.8 \times 10^{-4} \tag{2} \\
[\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \tag{3}
\end{align}$$
The next two equations are the so-called mass balance and charge balance equations:
$$\begin{align}
2[\ce{Ca^2+}] &= [\ce{F-}] + [\ce{HF}] \tag{4} \\
2[\ce{Ca^2+}] + [\ce{H3O+}] &= [\ce{F-}] + [\ce{OH-}] \tag{5}
\end{align}$$
Equation $(4)$ comes from the stoichiometry of $\ce{CaF2}$ dissociation. The total concentration of calcium-containing species, times two, must be equal to the total concentration of fluorine-containing species.
Equation $(5)$ comes from the fact that the system must be electrically neutral, i.e. the positive charges are the same as the negative charges. The concentration of calcium ions is weighted by 2 because it is doubly charged.
In this case, I just plugged it into Wolfram|Alpha, where the concentrations $[\ce{Ca^2+}]$, $[\ce{F-}]$, $[\ce{H3O+}]$, $[\ce{OH-}]$, and $[\ce{HF}]$ are represented by $p$, $q$, $r$, $s$, and $t$ respectively. The only solution that is physically sensible is the one where all the roots are positive and real. From this we find
$$[\ce{H3O+}] = r = 7.8 \times 10^{-8}$$
and hence $\mathbf{pH = 7.1}$.