solve a set in 'algerba of sets' way. 
*

*two question, need to solve each in 'algebra of sets' way
$$(A \cup B) - (C \cap  D) = (A - C) \cup (A - D)
    \cup (B - C) \cup (B - D)$$
$$(A \cup B) \cap  (C \cap  D)^\complement = (A
    \cap  C^\complement) \cup (A \cap  D^\complement)
    \cup (B \cap  C^\complement) \cup (B \cap 
    D^\complement)$$ becuase I know that for two sets, $A - B = A \cap B'$
$$(A \cup B) \cap  (C \cap  D)^\complement = A \cap 
    (C^\complement \cup D^\complement) \cup B \cap 
    (C^\complement \cup D^\complement)$$ distribute the right
side of the equation
$$(A \cup B) \cap  (C^\complement \cup D^\complement)
    = A \cap  (C^\complement \cup D^\complement) \cup B \cap  (C^\complement \cup D^\complement)$$ apply de
morgan on the left side
$$(A \cap  C^\complement) \cup (A \cap  D^\complement)
    \cup (B \cap  C^\complement) \cup (B \cap 
    D^\complement) = A \cap  (C^\complement \cup
    D^\complement) \cup B \cap  (C^\complement \cup
    D^\complement)$$ I don't know if it's legal? just how I see it, any
other way to distribute here?
$$A \cap  (C^\complement \cup D^\complement) \cup B
    \cap  (C^\complement \cup D^\complement) = A \cap 
    (C^\complement \cup D^\complement) \cup B \cap 
    (C^\complement \cup D^\complement)$$ distribute the left side

*$A \oplus B = A^\complement \oplus B^\complement$
$$(A - B) \cup (B - A) = (A - B)^\complement \cup (B -
    A)^\complement$$ i know the left side of the equation is legal for
me to do, but is the right side also?
$$(A \cap  B^\complement) \cup (B \cap  A^\complement)
    = (A \cap  B^\complement)^\complement  \cup (B \cap  A^\complement)^\complement$$ again I know that for 2 sets $A-B = A
    \cap B'$. is the right side of the equation legal for me to do?
can someone tell me how do I go from here?
 A: In the first problem, note that when you get to:
$$\begin{align} ...&= A \cap  (C^\complement \cup D^\complement) \cup B \cap  (C^\complement \cup D^\complement)\\ \\ & = (A \cap  C^\complement) \cup (A \cap  D^\complement)
    \cup (B \cap  C^\complement) \cup (B \cap 
    D^\complement)\end{align}\tag{1}$$
(You mention De Morgan's when moving from the left to the right side of the equation, but this equation holds by applying the distributive law.)
Note that you are virtually done. (All the work you do after obtaining $(1)$ is merely circular, taking you back to where you started.) 
Now from $(1)$, you need only apply the definition you suggested at the start: $$P - Q = P\cap Q^\complement$$
This gives you, as desired: $$(A \cap  C^\complement) \cup (A \cap  D^\complement)
    \cup (B \cap  C^\complement) \cup (B \cap 
    D^\complement) = (A - C) \cup (A - D) \cup (B-C) \cup (B - D)$$

$(2)$ From the start, the following is NOT valid: $$(A - B) \cup (B - A) = (A - B)^\complement \cup (B -
    A)^\complement$$
Rather, let's start off with what you know to be equivalent:
$$\begin{align} A \oplus B & = (A - B)\cup (B - A) \\ \\&= (A\cap B^\complement) \cup (B\cap A^\complement)\\ \\ & = \ldots\end{align}$$
