If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?

The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.

However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.

The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.

We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.

If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.

The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.

Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.