Let $A,B,C$ be events. Find an expression for the event "at least one of B and C occur, but A does not" Let $A,B,C$ be events. The event "$A$ and $B$ occur but $C$ does not" may be expressed as $A \cap B \cap C^c$. 
(a)  Find an expression for the event "at least one of B and C occur, but A does not"
(b)  Show that the probability of event in (a) is equal to
$\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(B\cap C)-\mathbb{P}(A\cap C)+\mathbb{P}(A\cap B \cap C)$

I claim that the answer to (a) is $(B \cup C)\cap A^c$. Can someone confirm or deny?
I have no idea how to proceed from here. From previous work, I have proven the following results:
$\mathbb{P}(A\setminus B)=\mathbb{P}(A)-\mathbb{P}(A \cap B)$
$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$
My main concern about (a) is what exactly they mean by "at least one" and the use of the operator "and".
 A: 
My main concern about (a) is what exactly they mean by "at least one" and the use of the operator "and".

"At least one of" a series of events means you have an inclusive union.  $A\cup B$ means the event of $A$ happening, or $B$ happening, or both happening.
"and" means the intersection.  $A\cap B$ means the event of $A$ and $B$ both happening.
So your answer is confirmed, $(B\cup C)\cap A^c$ and as a hint this is equal to $(B\cup C)\setminus A$.  Now apply your rules.
A: your answer to (a) is correct. "at least one of" means "either of, and possibly both". the "and" in "one of A and B" is not a logical operator, but is used as a list-former,e.g. "at least one of A,B,C and D"
the probability expression quoted seems to lack a term so i will insert it. we now have two main groups (parenthesized):
$$
(\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(B\cap C))-(\mathbb{P}(A\cap B)+\mathbb{P}(A\cap C)-\mathbb{P}(A\cap B \cap C))
$$
parses as (1): probability of at least one of B and C
$$
\mathbb{P}(B)+\mathbb{P}(C) -\mathbb{P}(B\cap C)
$$
the intersection is subtracted as otherwise it would be counted twice in the $\mathbb{P}(B)+\mathbb{P}(C) $ terms.
(2) removing the probability of A occurring
this mirrors the previous situation in (1), except that everything is intersected with A. in this case the $\mathbb{P}(A \cap B \cap C)$ is removed to correct for double counting. however, because this entire second component is being removed from the event, without the external parentheses this will occur with a positive sign, viz:
$$
\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(B\cap C)-\mathbb{P}(A\cap B)-\mathbb{P}(A\cap C)+\mathbb{P}(A\cap B \cap C)
$$
