Let $A = \mathbb{Z}$, $B = [−1, \pi]$, $C = (2, 7)$. List all elements of $A \cap (B^c \cap C)$. After working it out on a number line, I got: $\{4, 5, 6\}$.
As it stands, the expression contains the integers that do not belong to the set $B$ that cross into $C$. This would result in $4, 5, 6$. Number $7$ would not be included because of the soft brackets surrounding $C = (2, 7)$. Is this correct?

You're right. 
Here are the intermediate steps: $B^c=(-\infty, -1) \cup (\pi, \infty)$, so $B^c\cap C = (\pi, 7)$. 
And as you wrote, the only integers in that interval are $4,5,6$.
