Given $d>8$ boxes and $n$ balls. Consider event $A$=boxes numbered $1,2,3,4$ receive 0 balls.$B=3,4,5,6$ receive 0 balls, $C=5,6,7,8$ receive zero balls. What is $P(A \cup B \cup C)$?
Attempt:
Method 1:
$|(A\cup B \cup C)| = 8$ $\implies$ $P(A\cup B\cup C) = (\frac{d-8}{d})^n$
Method 2:
By inclusion exclusion:
$P(A\cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B) - P(B \cap C) - P(A \cap C) - P(A \cap B \cap C) = 3P(A) - 2P(A\cap B)$
where the last equality uses $P(A\cap B) = P(B \cap C)$ and $P (A \cap B \cap C) = 0$
But $(\frac{d-8}{d})^n \neq 3(\frac{d-4}{d})^n - 2(\frac{d-2}{d})^n$ for $n \geq 2$
As was already clarified in the comments, method $1$ is wrong since the union of events corresponding to sets is not the event corresponding to the union of the sets.
You also have the wrong sign on $P(A\cap B\cap C)$, but that seems to have been a typo, since you correctly cancelled it with $P(A\cap C)$.
In the last line, where you have $\left(\frac{d-2}d\right)^n$ it should say $\left(\frac{d-6}d\right)^n$, since $6$ boxes need to be empty for $A\cap B$ or $B\cap C$ to occur.