The probability you're asked to find is
\(\mathbb P(A\cup B)-\mathbb P(A\cap B)\)
where \(A\cup B\) is the event that either event occurs (A, B, or both), and \(A\cap B\) is the event that both events occur.
Recall that
\(\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)\)
which means
\(\mathbb P(A\cup B)-\mathbb P(A\cap B)=\mathbb P(A)+\mathbb P(B)-2\mathbb P(A\cap B)\)
You're told that \(\mathbb P(A)=0.5\), \(\mathbb P(B)=0.7\), and (if I'm reading the diagram correctly) \(\mathbb P(A\cap B)=0.3\).
So,
\(\mathbb P(A\cup B)-\mathbb P(A\cap B)=0.5+0.7-2\times0.3=0.6\)
Another way of seeing this is that the event A consists of the regions "A not B" and "A and B". So the probability that "A not B" occurs is
\(\mathbb P(A\setminus B)=\mathbb P(A)-\mathbb P(A\cap B)=0.5-0.3=0.2\)
Similarly, B consists of "B not A" and "A and B", so you have
\(\mathbb P(B\setminus A)=\mathbb P(B)-\mathbb P(A\cap B)=0.7-0.3=0.4\)
So the probability that A or B, but not both, occur is
\(\mathbb P(A\setminus B)+\mathbb P(B\setminus A)=0.2+0.4=0.6\)
