Well, for the first one, we will assume ac > bc (the opposite). Since c>0, we can divide both sides by c and the greater-than does not become a less than.

which gives us a > b . However, this goes against the previous statement that a < b, which is a contradiction. Therefore, ac < bc.

For the second one it would have to be ab > 0, because two negatives multiplied always make a positive. Not entirely sure how to prove that, though. Seems so elementary...

Hope this helped a bit...