How can we know for use that $A$ and $B$ are mutually exclusive under conditonal probability? Reading that
"Sample question 4 (Conditional Probability): Given that $P(A) = 0.20, P(B) = 0.60, P(B|A) = 0.50,$ what is $(B∩A’)$ ? $(A’∩B’)$?" 
from datasciencecentral.com
When we are just provided with the text without picture/table. The only picture from the text in the head is:
           B         B'          Marginal                           
A       (.2*.5=.1)               0.2                  
A'                                                                 
Marginal  0.6

Then: 
           B          B'          Marginal                           
A       (.2*.5=.1)    (.1)        0.2                  
A'         (.5)                                                        
Marginal   0.6

in order to fill out the rest of table, we need to know that "A’ and B’ are mutually exclusive. "
Then the questions is: when we are dealing with Conditional Probability $P (A'  \cap  B')$ are always $0$ ? 
While the solution is given as :
           B          B'          Marginal                           
A       (.2*.5=.1)    (.1)        0.2                  
A'         (.5)       (0 ?)       (0.5 ?)                                           
Marginal   0.6        (0.1 ?)     (1.0 ? especially this one ) 

Can someone please help me understand where the parts with the question mark (above) comes form ? Thank you in advance.

 A: First off: No, these events cannot be mutually exclusive, since
$$P(A|B)\stackrel{\mathop{def}}{=}\frac{P(A∩B)}{P(B)}\ne0\ \implies \boxed{P(A∩B)\ne0.}$$
Although the rest of your question also needs some attention since I feel there is some misunderstanding on how things work.
Since every event partition (mutually exclusive & covering entirely) the sample space in two pieces we must have:
$$P(A\cup A')=P(A)+P(A')=P(B\cup B')=P(B)+P(B')=1$$
In your case, $P(A)$ is the "marginal" probability of $A$. The idea is that regardless of how $B$ behaves, $A$ should be consistent: It either happens or not. In your case, you already found both $P(A)$ and $P(B)$,
          B       B'      Marginal
A          0.1     0.1      0.2
A'         0.5              
Marginal   0.6              

so you can just assign $P(A')$ and $P(B')$ using these:
          B       B'      Marginal
A          0.1     0.1      0.2
A'         0.5              0.8
Marginal   0.6     0.4      

Notice that the sum of the two marginal probabilities do not add up to $1$, on the contrary they add up to $2$ since it is the addition of the total probabilities of two events. This carries to cases with more events since $\sum_{i=1}^N P(A_i\cup A_i')=\sum_{i=1}^N 1=N$.
In the last step, you do the same algorithm. The rows and columns should add up to the last element (i.e. the "marginal" probabilities).
          B       B'      Marginal
A          0.1     0.1      0.2
A'         0.5     0.3      0.8
Marginal   0.6     0.4      2.

You can fill it using the row and since the system is consistent the column is automatically satisfied (you can check). Though since the variables are not independent I do not advise heavy usage of tables (since they make people recall their earlier, simpler days with inherent assumption of independence).
P.S. It really is criminal that SE doesn't support tables. Like, really?
