1) In how many ways can $12$ children occupy the six banks of two seats on a Ferris wheel? (consider important the order of two children sitting in a given bank). Resp.: 7680
2) What would be the answer from the previous exercise if the order of two children sitting in a given bank was not of interest? Resp.:120
Usually questions like this mean that the author only cares about the order of the people sitting; i.e. you need only care, for an individual, who sits besides them on either side. The rationale is obvious given this is a Ferris Wheel and everyone is spinning, although one might argue that the actual seat matters.
So essentially this boils down to fixing one person and then seeing how the others can be arranged around the person. Since there are $6$ banks, this falls to $5!=120$, which is the answer to the second part.
For the first part, you also need to care about who sits on the left and who on the right, so for each bank, you have two possibilities. So your answer is obtained by multiplying $120$ by $2^6$ which is exactly $7680$.