There are $18$ courses that are silver or bronze, with twice as many bronze as silver. So there are $2$ gold, $6$ silver, and $12$ bronze.
For the first question, there are $8$ choices.
For the second question, the place for the first round can be chosen in $12$ ways. For each of these choices, the place for the second round can be chosen in $6$ ways. And for each way of choosing where to play the first two rounds, there are $2$ choices for the location of the third round.
Thus the total number of choices is $(12)(6)(2)$.
Remark: We did not worry about whether "permutation" or "combination" was involved. We just solved the problem.
The only thing that was used is what is sometimes called the Multiplication Principle. If one can do Task $1$ in $a$ ways, and for each of thse ways we can do Task $2$ in $b$ ways, then we can do Tasks $1$ and $2$ in $ab$ ways. One should not memorize this as a rule, it should just be obvious. If Alicia has $a$ children, and each child has $b$ children, then Alicia ends up with $ab$ grandchildren. And if each grandchild has $c$ children, Alicia ends up with $abc$ great-grandchildren. And so on.
I will take the risk of confusing you, and say that the problem has elements of both permutations and combinations!
We want to count the number of sequences $(A,B,C)$, where $A$ is a bronze golf course, $B$ is a silver, and $C$ is a gold. Sequences sound like permutations.
But to put it another way, we want to choose a collection of three golf courses, with some restrictions: one of the choices has to be gold, another silver, another bronze. "Choose" sounds like combinations.