I think you are right.
See page 133:
In free logic, the quantifiers have existential import, but constants do not.
The formula $(Fa ⊃ (∃x)Fx)$ is not valid.
Why so ? Because we have to consider a "world" where $Fa$ is true but $(∃x)Fx$ is false, because the constant $a$ has not existential import.
We can check it verifying that the proof tree deos not close:
1) $\lnot (Fa ⊃ (∃x)Fx)$
3) $\lnot (∃x)Fx$
4) $(x) \lnot Fx$.
But we cannot use (FUI) to instantiate 4) with $\lnot Fa$ because we do not have $(∃x)(x = a)$.
$((∃x)(a ≠ x) \& Fa) ⊃ (∃y)Fy$.
Counterexample : World $n$ is empty, i.e. $Q(n)=\emptyset$ but $a ∈ D(n)$.
It seems to me that it does not work; in an empty domain, every existentially quantified formula $(∃x)Fx$ is false (there are no elements in $Q$ to satisfy it) and every universally quantified formula $(x)Fx$ is true (there are no elements in $Q$ to falsify it).
Thus, $(x)(x=a)$ is true because there are no objects in the domain of quantification $D$, and so $\lnot (x)(a=x)$, i.e. $(∃x) \lnot (a = x)$ is false.
Due to $a ∈ D(n)$, we have: $Fa [n] = 1$ but: $(∃x)(a ≠ x) [n] = 0$. $Q(n)$ is empty, and thus: $(∃y)Fy [n] = 0$.
Conclusion: $((0 \& 1) ⊃ 0) [n] = 1$.
We may "fix" it considering a one-element domain $Q$ with $b \in Q$ and not $Fb$.
In this case, we still have: $Fa [n] = 1$, due to $a ∈ D(n)$, but: $(∃x)(a ≠ x) [n] = 1$. Finally, not $Fb$, and thus: $(∃y)Fy [n] = 0$.
Again, we can check that the proof tree does not close:
1) $\lnot [ \ ((∃x)(a ≠ x) \& Fa) ⊃ (∃y)Fy \ ]$
2) $(∃x)(a ≠ x) \& Fa$
3) $\lnot (∃y)Fy$
4) $(∃x)(a ≠ x)$ --- from 2)
5) $Fa$ --- from 2)
6) $(a \ne b)$ --- from 4 by (FEI)
7) $(∃z)(z = b)$ --- from 4 by (FEI)
8) $(y) \lnot Fy$ --- from 3)
and we are stuck; we can apply (FUI) to 8) because we have : $(∃z) (z = b)$ to get $\lnot Fb$, but this does not close with 5).