Why aren't these events mutually exclusive? A parallel system functions whenever at least one
of its components works. Consider a parallel system
of three components, and suppose that each component
works independently with probability $0.5$.
Find the conditional probability that component 1
works given that the system is functioning.
lets say $A_i =$ the event that the $i$th component works. 
If they're not mutually exclusive, what would $A_1 \cap A_2$ be?? 
 A: If $A_1$, $A_2$ are independent, then
$$\Pr(A_1\cap A_2) = \Pr(A_a)\cdot\Pr(A_2) = (0.5)(0.5) = 0.25 \ne 0.$$
If $A_1$ and $A_2$ were mutually exclusive, then that probability would be $0$.
The question in your title asks why they are not mutually exclusive.  The above should answer that.
In the body of your question you ask what $A_1\cap A_2$ is.  It's just the event that the first two components both work.
A: For the conditional probability, you can see that there are 8 equally likely outcomes possible for the functionality of the components, only one of which results in the system not working.
Of the seven working systems, three have component #1 not working and four have component #1 working.
So, the condition probability that component #1 is working, given that the whole system is working is $4/7$.
(This whole setup is just like a three coin flip experiment, where you are asking "If a friend flipped a coin three times and told you at least one head appeared, what is the probability a heads appeared on your friend's first flip?"
