Probability of attending cinema and theater. Problem: 
It is researched that 60% of people in city goes to cinema on daily basis, 40% of people goes to theater on daily basis. It is also known that 20% simultaneously goes to both theater and cinema.

What is



Probability that chosen person does not attend both.
Probability that chosen person attends theater OR cinema.
Probability that person, which attends cinema will
attend theater too.
Probability that person, which attends theater will
attend cinema too.
Probability that person, which doesn't attend theater
will attend cinema.
Probability that person, which doesn't attend cinema
will attend theater.


So far I have got:


A-person attends cinema
B-person attends theater


1)$P(\overline{A}\cap\overline{B})=P(\overline{A})\cdot P(\overline{B})=0.4\cdot 0.6=0.24$

2)$P(A \cup B)=P(A)+P(B)-P(A \cap A)=0.6+0.4-0.24=0.76$

I have no clear understanding of how to solve rest of questions.

For example,  in 3rd questions I guess that I need to find $P(A \cap B)$, which gives exactly the same as 4th questions $P(B \cap A)$, which seems unlikely for me.

And I guess that 5th and 6th would be:

5) $P(\overline{B}\cap A)=P(\overline{B})\cdot P(A)=0.6\cdot
   0.6=0.36$

6) $P(\overline{A}\cap B)=P(\overline{A})\cdot P(B)=0.4\cdot 0.4=0.16$

Is there a something I am not getting here?
First a note to your solution of item (2), it should be $P(A)+P(B)-P(A\cap B)$. Regarding item (3)-(4), you are in fact not looking for the joint probability, but the conditional probability i.e. $P(B|A)$ and $P(A|B)$ respectively, where the conditional probability is defined as usual:
\begin{equation}
P(X|Y) = \frac{P(X\cap Y)}{P(Y)}.
\end{equation}
The solutions to items (5)-(6) are found using the conditional probability as well. That is, by $P(A|\bar{B})$ and $P(B|\bar{A})$ respectively.

Hope that helps.
On your answer (1), you are implying that the two events are independent which does not seem to be the case.
I would have use the De Morgan's Law to solve it:
$$ \Pr \big( ( A \cup B ) ^ { \mathrm c } \big) = 1 - \Pr ( A \cup B ) = 0.2 $$
$$ \Pr \big( ( A \cup B ) ^ { \mathrm c } \big) = \Pr ( A ^ { \mathrm c } \cap B ^ { \mathrm c } ) $$