Does A∪C=B∪C imply A = B? What about intersection instead of union? if A, B and C are sets; 
for (u = union, n = intersection)

$A \cup C = B \cup C$, would $A = B$ ?

Similar for 

$A \cap C = B \cap C$, would $A = B$ ? 

For my conclusion I have got for the first one, $A \neq B$ because $A$ can contain values differently from $B$ and for the second one $A$ does equal to $B$. Is this correct? 
 A: Your answer is correct in the first case. But to be complete you need to give an example. Consider the sets
$A=\{1,2\},$ $B=\{2,3\}$ and $C=\{1,2,3\}.$ It is $A\cup B=A\cup C=C=\{1,2,3\}.$ However, $A\ne B.$
For the second case, consider $A=\{1,2\},$ $B=\{2,3\}$ and $C=\{2\}.$ It is $A\cap B=A\cap C=C=\{2\}.$ But, $A\ne B.$
A: That depends on what quantification you do on $C$. If the premise is true for all sets $C$ then your conclusion holds, not if it's a specific unfortunate set.
To see why this is the case in the first interpretation you can choose $C$ to be a fortunate set. If we have $C=A\cap B$ in the first equation and $C=A\cup B$ in the second:
$A = A \cup (A \cap B) = B \cup (A \cap B) = B$
$A = A \cap (A \cup B) = B \cap (A \cup B) = B$
However if the premises don't hold for any set $C$ we may be unfortunate with the value $C$ has. In the first equation we could have that $C = A\cup B$ for the first equation and $C = A \cap B$ for the second and we would have regardles of $A$ and $B$ (which could be unequal):
$A \cup C = A \cup (A \cup B) = A\cup B = B \cup (A\cup B) = B \cup C$
$A \cap C = A \cap (A \cap B) = A \cap B = B \cap (A \cap B) = B \cap C$
A: Hint
$1)$ Look what happen if $A\neq B$ and $A,B\subset C$.
$2)$ Look what happen if $A\neq B$ and $A,B\supset C$.
