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?
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):
$A \cup C = A \cup (A \cup B) = A\cup B = B \cup (A\cup B) = B \cup C$
$A \cap C = A \cap (A \cap B) = A \cap B = B \cap (A \cap B) = B \cap C$