We distribute the integers $1, 2, \ldots, 8$ into two sets $A$ and $B$, then we denote $P_{A}$ as the product of all elements in $A$ and $P_{B}$ as the product of all elements in $B$.
What are the minimum and maximum values that the sum $P_{A} + P_{B}$ can take?
Note: if a set $E$ is empty, we consider the product of its elements to be equal to 1.
Let $A$ and $B$ be two disjoint sets whose union is equal to the set $E=\{1, \ldots, 8\}$.
First, let's maximize the sum $P_{A}+P_{B}$. Without loss of generality, we can assume that $P_{A} \leqslant P_{B}$. Then, if $A$ contains an integer $k \geqslant 2$, we set $A^{\prime}=A \backslash\{k\}$ and $B^{\prime}=$ $B \cup\{k\}$. Then

$$
P_{A^{\prime}}+P_{B^{\prime}} \geqslant P_{B^{\prime}}=k P_{B} \geqslant 2 P_{B} \geqslant P_{A}+P_{B} .
$$

Thus, when the sum $P_{A}+P_{B}$ reaches its maximum value, we know that $A \subseteq\{1\}$, so $P_{A}=1$ and $P_{B}=8!$, hence $P_{A}+P_{B}=8!+1=40321$.
Now, let's minimize the sum $P_{A}+P_{B}$. The arithmetic-geometric mean inequality indicates that $P_{A}+P_{B} \geqslant 2 c$, where we have set $c=\sqrt{P_{A} P_{B}}=\sqrt{8!}$. Since $401^{2}=160801401$, and thus $P_{A}+P_{B} \geqslant\lceil 2 c\rceil \geqslant 402$.
A first instinct is therefore to find two sets $A$ and $B$ such that $P_{A}+P_{B}=402$. We then make some preliminary remarks for this search. First, knowing which set contains the integer 1 makes no difference. Then we prove that one of the two sets contains the integer 6 and the other contains the integers 2 and 3. Indeed, let $X$ be the set that contains 6 and $Y$ the one that does not contain 6:
$\triangleright$ since $P_{Y} \equiv 402-P_{X} \equiv 0(\bmod 6)$, it follows that $Y$ contains the integer 3 and an even integer;
$\triangleright$ if it is $X$ that contains the integer 2, then $P_{Y} \equiv 402-P_{X} \equiv 2(\bmod 4)$, so $X$ contains an even integer but not divisible by 4, which is impossible.
By removing the integers $1,2,3$ and 6 from the sets $A$ and $B$, we are thus reduced to finding a partition of the set $\hat{E}=\{4,5,7,8\}$ into two subsets $\hat{A}$ and $\hat{B}$ such that $P_{\hat{A}}+$ $P_{\hat{B}}=402 / 6=67$. Suppose, without loss of generality, that $\hat{A}$ contains the integer 4. Then $P_{\hat{B}} \equiv 67-P_{\hat{A}} \equiv 1(\bmod 2)$, so $\hat{A}$ also contains the integer 8. Furthermore, if $\hat{A}$ also contains one of the two integers 5 or 7, then $P_{\hat{A}}+P_{\hat{B}} \geqslant P_{\hat{A}} \geqslant 4 \times 5 \times 8 \geqslant 160>67$. Therefore, we must necessarily have $\hat{A}=\{4,8\}$ and $\hat{B}=\{5,7\}$, and we are happy to verify that, in this case, we indeed have $P_{\hat{A}}+P_{\hat{B}}=32+35=67$.
In conclusion, the maximum value of $P_{A}+P_{B}$ is equal to $P_{\emptyset}+P_{E}=8!+1=40321$, and the minimum value of $P_{A}+P_{B}$ is equal to $P_{\{2,3,5,7\}}+P_{\{1,4,6,8\}}=402$.