Let $\Phi_e$ denote the $e^\text{th}$ Turing machine. One form of the Halting problem is that the set $\{e : \Phi_e(e) \text{ halts }\}$ is not computable. It is good exercise to show that this Halting Problem is equivalent to whatever form of the halting problem you are using.
If fixed finitely many Turing Machines $F = \{e_0, e_1, e_2, ..., e_n\}$, then there exists finite subsets $F_0, F_1 \subseteq F$ such that for all $e \in f_0$, $\Phi_e(e)$ does not halt and for all $e \in F_1$, $\Phi_e(e)$ converges. Then you can define a computable function
$\Psi(n) = \begin{cases}
1 & \quad e \in F_1 \\
0 & \quad e \in F_0 \\
0 & \quad \text{ otherwise }
\end{cases}$
Since $F, F_0, F_1$ are all finite, you can make a Turing Machine that compute $\Psi$. Hence $\Psi$ is a Turing Machine (or computable function) that tells the answer to the Halting Problem for the particular finite set $\{e_0, e_1, ..., e_n\}$.
Note that this process is not uniform. If you fixed $F$, there exists a Turing machine that $\Psi_F$ that tell you the answer for $F$. However, there is not computable function taking input $F$, will give you computably $\Psi_F$. This is because given $F$, there is not computable procedure to tell you what $F_0$ and $F_1$ are. In the case that you fixed a particular $F$, it was good enough just to know that there exists finite sets $F_0$ and $F_1$ that works for this particular $F$.
