1. (12 P 6) / 6 = 12!/6!*6 = (12*11*10*9*8*7) / 6 = 110,880
2. (12 P 6) = 12!/6! = (12*11*10*9*8*7) = 665,280
For a bench (easier to describe): You want to sit one of the 12 in the first spot, then one of the remaining 11 in the second spot, one of the remaining 10 in the third spot, ..., and one of the remaining 7 in the sixth spot. Order matters here so the quick way is to use a permutation (12 P 6) and you get your answer.
For a round table you use the same idea as above. Choose an arbitrary starting point and you can put any of the 12 people there. Go around the table and end with the final spot, where you can put any of the remaining 7 people. Again, order matters so a permutation is used (12 P 6). The difference here is that you have a round table (note that it doesn't have to be 'round', it can be square or rectangular or whatever so long as it loops around and back to the beginning). Imagine I choose people A, B, C, D, E, and F and sit them around a table so that person A is at the 12 o'clock position and B through F are in that order clockwise. That's 1 of the 665,280 possibilities. Now imagine I shift everyone one seat clockwise but keep the order. That's 2 of the 665,280 possibilities. By repeating this a few more times, you come to realize that 6 of the 665,280 orders come from seating A-F in one order around the table. Since there are 6 ways to position the first person (6 seats), you only need to consider 1 out of the 6 ways because we consider them to be the same.
I know their physical positions are different so you might be wondering why this is different from the bench. Well just remember that when people sit on a straight bench, the first and last people are not next to each other. So depending on where you are sitting, you may be next to 1 person or 2 people. At a round table, you are always between two people.
So since 1 out of 6 are unique, you take 665,280 / 6 and get 110,880.
