when 2 events A and B are indep , the occurrence of one event A doesnt affect the occurrence of the other event B and P(A intersection B) = P(A)*P(B) 
Again, when A is indep of the occurrence of the other event B it goes without saying :A is indep of the nonoccurrence of B.i.e. A is indep of Bcap Hence P(A intersection Bcap = P(A)*P(Bcap) 
These are the 2 rules we are going to make use of here. 
P(A|B) = P(A intersection B) /P(A) = P(A)*P(B) /P(A) = P(B) = 0.4 

P(A \cup B) = P(A intersecton B cap)/ P(Bcap) = P(A)*P(B cap)/ P(Bcap) = P(A) = 0.6 

third subdiv is the same as the first one 
If you've meant P(B\cup A ) we ' ll use the same method specified above. 
= P(B intersecton A cap)/ P(Acap) = P(B)*P(A cap)/ P(Acap) = P(B) = 0.4 

b) A & B are dependent.so we CAN NOT use the rule P(A intersection B) = P(A)*P(B) 

P(A/B) = P(AB)/P(B) = 0.3 (given) 
so P(AB) / 0.4 = 0.3 
P(AB) = 0.4*0.3 = 0.12 
P(B/A ) = P(AB)/P(A) = 0.12 /0.6 = 0.2 
P(A \cap B) = P(A intersection B cap) / P(Bcap) ..............Denote this by (M) 
To find P(A intersection B cap) , consider (A intersection B cap ) in the Venn Diagram. 
[That is you draw two circles interscting one another inside a rectangle.the 2 cicles correspond to the sets A & B . The rectangle represents the sample space] 
You may notice (A intersection B cap) is A - AB 
So P(A intersection B cap) = P( A - AB) = P(A) - P(AB) = 0.6 - 0.12 = 0.48 
in the expression denoted by (M) we have obtained the numerator. 
To get the denominator take P(Bcap) = P( non- occurrence of B) = 1- P(B) = 1 - 0.4 = 0.6 
So (M) becomes 0.48/0.6 =4/5 = 0.8