(a) The fact that $P(A \cup B)=P(A)+P(B)-P(A\cap B)$ is useful, and the fact that in this case, by independence, we have $P(A \cap B)=P(A)P(B)$ Let $P(B)=y$. Then $0.4=0.2+y -(0.2)y.$ The above equation is linear in $y$, and easy to solve. We get $y=0.25$.
(b) A peculiar question! This asks for $P(A \cup B)$, but we have been told that this is $0.4$.
(c) You are probably expected to use the defining formula for conditional probability, namely that $P(C|D)=\frac{P(C\cap D)}{P(D)}. \qquad(\ast)$ In our problem, $C$ is the event $A$, and $D$ is the event "exactly one of $A$ and $B$."
First we find $P(D)$. The probability that exactly one of $A$ and $B$ occurs is $P(A \cup B)$ minus the probability that $A$ and $B$ both occur. But since $P(A)=0.2$ and $P(B)=0.25$, by independence $P(A \cap B)=(0.2)(0.25)=0.05$. It follows that the probability that exactly one of $A$ and $B$ occurs is $0.4-0.05$, which is $0.35$.
Now we need $P(C \cap D)$, the probability that $A$ occurs and exactly one of $A$ and $B$ occurs. This is a long-winded way of saying that $A$ occurs and $B$ does not. We have $P(A)=0.2$. By our earlier computation, $P(B)=0.25$, so $P(B^c)$ (the probability $B$ does not occur) is $0.75$. By independence, $P(A \cap B^c)=(0.2)(0.75)=0.15$.
Substituting in $(\ast)$, we find that the probability that $A$ occurs given that exactly one of $A$ and $B$ occurs is $\dfrac{0.15}{0.35}$, which looks better as the fraction $2/7$.