Added. It seems to me that you need to back up and start slower, develop some feel for these operations and their meanings; right now, it seems like you are just trying to shuffle symbols...
You should become familiar with some of the basic properties of the union, intersection and complement: for all sets $X$, $Y$, and $Z$:
- Commutativity of union: $X\cup Y = Y\cup X$.
- Commutativity of intersection: $X\cap Y=Y\cap X$.
- Associativity of union: $(X\cup Y)\cup Z = X\cup(Y\cup Z)$.
- Associativity of intersection: $(X\cap Y)\cap Z = X\cap (Y\cap Z)$.
- Distributivity of intersection over union: $X\cap(Y\cup Z) = (X\cap Y)\cup(X\cap Z)$.
- Distributivity of union over intersection: $X\cup(Y\cap Z)=(X\cup Y)\cap (X\cup Z)$.
- Idempotency of union: $X\cup X = X$.
- Idempotency of intersection: $X\cap X = X$.
- Empty set is the neutral element for union: $X\cup\emptyset = X$.
- Empty set is a zero element for intersection: $X\cap \emptyset=\emptyset$.
- Double complementation: (X')' = X.
- Complements and unions: X\cup X' = \text{everything}.
- Complements and intersections: X\cap X' = \emptyset.
- De Morgan's Laws: (X\cap Y)' = X'\cup Y' and (X\cup Y)' = X'\cap Y'.
Try proving the following:
- $X\subseteq Y$ if and only if X\cap Y'=\emptyset.
- $X\subseteq Y$ if and only if X'\cup Y =\text{everything}.
- $X\subseteq Y$ if and only if (X\cap Y')\subseteq X'.
- $X\subseteq Y$ if and only if (X\cap Y')\subseteq Y.
- $X\subseteq Y$ if and only if (X\cap Y')\subseteq (Z\cap Z').
You have to be careful. Just because two expressions are equal does not necessarily mean that there is some standard rewriting method that transforms the first into the second. Just because $2\times 3 = 1+5$, that does not mean that you have some standard way of taking $2\times 3$ and having it "become" $1+5$; it's just that the two expressions yield the same final answer.
Here, though, you do have some basic rules, specifically distributivity. For any sets $X$, $Y$, and $Z$, we have that $\cup$ distributes over $\cap$, and $\cap$ distributes over $\cup$ (just like, with numbers, $\times$ distributes over $+$). Explicitly, $X\cap (Y\cup Z) = (X\cap Y)\cup(X\cap Z)\qquad\text{and}\qquad X\cup(Y\cap Z) = (X\cup Y)\cap(X\cup Z).$ This just says: the things that are both in $X$ and (either in $Y$ or $Z$) are the same as the things that are either in both $X$ and $Y$, or in both $X$ and $Z$; and the things that are either in $X$ or in both $Y$ and $Z$ are the same as the things that are both in either $X$ or $Y$, or in either $X$ or $Z$.
So, take (A'\cup B')\cap A\cap C. This is the same (since $(X\cap Y)\cap Z = X\cap (Y\cap Z)$; intersection is associative) as \Bigl( (A'\cup B')\cap A\Bigr)\cap C. Distributing the first $\cap$ over the $\cup$, we have that this is the same as \Bigl( (A'\cap A)\cup (B'\cap A)\Bigr)\cap C. Distributing the last $\cap$ over the $\cup$, we get that this is the same as \Bigl((A'\cap A)\cap C\Bigr) \cup \Bigl( (B'\cap A)\cap C\Bigr). Now, look at A'\cap A\cap C: since A' is the complement of $A$, then A'\cap A = \emptyset, so A'\cap A\cap C=\emptyset. And since for any set $X$, $\emptyset\cup X = X$, then the entire thing is just the same as the second set, namely (B'\cap A)\cap C. And since $X\cap Y = Y\cap X$, then this is the same as A\cap B'\cap C.
So, no it should not be A\cup B'\cap C.
No, it is not true that the parentheses have no effect in general. Both $\cup$ and $\cap$ are associative, so that $(X\cup Y)\cup Z$ is the same as $X\cup(Y\cup Z)$, and so we may write it as $X\cup Y\cup Z$ (exactly like $(2+3)+4 = 2+(3+4) = 2+3+4$). But when you have both unions and intersections, parenthesis start to matter because $(X\cup Y)\cap Z$ is not necessarily the same thing as $X\cup(Y\cap Z)$, exactly like $(2+3)\times 4$ is not the same as $2+(3\times 4)$.