How can you use ICE tables to solve multiple coupled equilibria?
The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.

Here is the combined ICE table:

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &amp;[\ce{Ca^2+}]  &amp; [\ce{F-}]  &amp; [\ce{H+}]&amp;[\ce{HF}]   \\
\hline
I &amp;  \pu{2.1e−4} &amp; \pu{4.2e−4} &amp; \text{N/A} &amp; 0 \\
\hline
C &amp;  +x  &amp; +2x-y &amp; \text{N/A} &amp; +y  \\
\hline
E &amp;  \pu{2.1e−4}+x &amp; \pu{4.2e−4}+2x-y &amp; 10^{-3.00} &amp; +y \\
\hline
\end{array}
$$

Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.

What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.

Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &amp;[\ce{Ca^2+}]  &amp; [\ce{F-}]  &amp; [\ce{H+}]&amp;[\ce{HF}]   \\
\hline
I &amp;  0 &amp; 0 &amp; \text{N/A} &amp; 0 \\
\hline
C &amp;  +p  &amp; +2p-q &amp; \text{N/A} &amp; +q  \\
\hline
E &amp;  +p &amp; +2p-q &amp; 10^{-3.00} &amp; +q \\
\hline
\end{array}
$$