Source: Internet newsgroup rec.puzzles.

• 5*log5 (where the base of the log is 2), which is around 11.
• 10 (compare each weight to each of the others).
• 8. A good description of this comes from Craig Gentry:

     First, use 3 weighings to sort 3 of the weights.  Next, weigh the
   median weight against each of the two remaining weights.  If the median
   weight is heavier than both or lighter than both, then use three
   weighings to sort the three unsorted weights.  If the median weight is
   heavier than one but lighter than the other, then use two weighings to
   sort each of the two unsorted pairs.
  

• 7. Craig Gentry took my suggestion of looking for a shorter solution and found a good one. First mine and then his solutions follow:
  Ken's solution:
  Weigh two weights against each other. Weigh two others against each other, and weigh the heaviest against each other. Label the weights such that A>B, A>C>D, and E is unknown. These first three weighings reduced the possible orderings from 120 to 15, and the following weighings continue to divide the possibilities in half. The next two weighings find the position of E in ACD, and the final two weighings find the position of B in the last four positions.

  E>C, E>A, B>C        EABCD
  E>C, E>A, C>B, B>D   EACBD
  E>C, E>A, C>B, D>B   EACDB
  E>C, A>E, B>C, B>E   ABECD
  E>C, A>E, B>C, E>B   AEBCD
  E>C, A>E, C>B, B>D   AECBD
  E>C, A>E, C>B, D>B   AECDB
  C>E, E>D, B>E, B>C   ABCED
  C>E, E>D, B>E, C>B   ACBED
  C>E, E>D, E>B, B>D   ACEBD
  C>E, E>D, E>B, D>B   ACEDB
  C>E, D>E, B>D, B>C   ABCDE
  C>E, D>E, B>D, C>B   ACBDE
  C>E, D>E, D>B, B>E   ACDBE
  C>E, D>E, D>B, E>B   ACDEB
  

  
  Craig's solution:
  

  Hmm, I knew that a process using 7 weighings was theoretically possible, since
  7 weighings can distinguish among 2^7 = 128 possibilities, more than the 120
  possible permutations.  However, I didn't really expect for such a process
  to exist.  I guess I should have looked more closely.
  
  Call the weights A, B, C, D, and E.  Weigh A against B.  WLOG, suppose A is
  lighter than B.  It is clear that our next weighing cannot include either A
  or B, because if, say, A is lighter than C, then there remain 2 (of 6)
  possible orderings of A, B, and C, leaving 40 (120*2/6) possible orderings
  of A, B, C, D, and E - more than we can handle with 5 weighings.  Therefore,
  weigh C against D. WLOG, suppose C is lighter than D.  Actually, for the
  sake of clarity, let me at this point revert to an outline form:
  
  1) A weighs in lighter than B.
  2) C weighs in lighter than D.
  3) WLOG, A weighs in lighter than C.
  4) Weigh C against E.
          5) If C is lighter, weigh D against E.
                  6) If D is lighter, weigh B against D
                          7) If B is lighter, weigh B against C.
                                  a) If B is lighter, ABCDE.
                                  b) If C is lighter, ACBDE.
                          7) If D is lighter, weigh B against E.
                                  a) If B is lighter, ACDBE.
                                  b) If E is lighter, ACDEB.
                  6) If E is lighter, weigh B against E.
                          7) If B is lighter, weigh B against C.
                                  a) If B is lighter, ABCED.
                                  b) If C is lighter, ACBED.
                          7) If E is lighter, weigh B against D.
                                  a) If B is lighter, ACEBD.
                                  b) If D is lighter, ACEDB.
          5) If E is lighter, weigh A against E.
                  6) If A is lighter, weigh B against C.
                          7) If B is lighter, weigh B against E.
                                  a) If B is lighter, ABECD.
                                  b) If E is lighter, AEBCD.
                          7) If C is lighter, weigh B against D.
                                  a) If B is lighter, AECBD.
                                  b) If D is lighter, AECDB.
                  6) If E is lighter, weigh B against D.
                          7) If B is lighter, weigh B against C.
                                  a) If B is lighter, EABCD.
                                  b) If C is lighter, EACBD.
                          7) If D is lighter, EACDB.
  
  Given the complexity of the above process, it seems impossible to determine
  a formula for how many weighings are necessary to sort n weights.  A decent
  upper bound is (n(k+1) - 2^(k+1) + 1), where 2^k < n <= 2^(k+1).  Or to make
  it more intuitive, 2^k(k-1) + 1 is an upper bound for 2^k weights.  We can
  get this upper bound inductively by observing that if n-1 of n weights are
  already sorted, it takes at most k+1 more weighings to sort the nth weight.
  But, of course, this upper bound is far from perfect, as it equals 8 for 5
  weights.  So, we can reduce this upper bound to n(k+1) - 2^(k+1) for n
  weights or 2^k(k-1) for 2^k weights, but this bound is probably still not
  very tight.
  
  -Craig Gentry