Finding Unions and intersections given two probabilities. I am currently trying to find the unions and probabilities given:
A = .2 and B = .6. P($A\cap B$) = .12
And am looking to find the following:
P($A^c\cap B$) For this I just did $.8*.6$ which gives .48.
P($A\cup B$) Here I apply P(A) + P(B) - P($A\cap B$) -> (.2+.6)-(.12) = .68
P($(A\cup B)^c$) Here I just give the opposite of the previous answer -> .32
Did I approach these probabilities correctly? I was unsure of the formulas to use in order to solve these probabilities. 

Because $\Pr(A\cap B)$ is in this case equal to $\Pr(A)\Pr(B)$, the events $A$ and $B$ are independent, so the answer you got is correct.  
However, you would not necessarily be viewed as having written a correct solution. If for example $\Pr(A\cap B)=0.1$, multiplying $0.8$ by $0.6$ would yield the wrong answer.  
I suggest using an argument like this. We have
$$\Pr(A^c\cap B)+\Pr(A\cap B)=\Pr(B).$$
We know two of the terms, so we can find the third, which is $0.6-0.12$.
Your other two arguments are correct.
