1. f is Theta(n) means there exists constants c, C, and N such that
      cn <= f(n) <= Cn for all n>=N.
   g is O(n) means that there exists constants C', N' such that
            g(n) <= C'n for all n>= N'.
   Assume g(n) >= 0 for all n (needed for the left-hand inequality below.
   I should have made that clear in the problem statement). We then have 
   that
              cn <= f(n) + g(n) <= Cn + C'n for all n >= max{N,N'}.

   So: select d = c and D = max{C,C'} (or D=C+C') 
   and M = max{N,N'} and we see that

              dn <= f(n) + g(n) <= Dn for all n>=M.

   Thus  f(n) + g(n) is Theta(n).

2.  Assume that when we break up a list of odd length n, we do it
    floor{n/2} elements on the left and ceiling{n/2} on the right.
    Then: make sure that at each stage, the largest value is on the right, 
    and the second largest value is on the left.   Lots of ways to do
    this.  For example: 

    45678     01239
   67  458    23  019
       45 8       01  9
  
   Solution: 6 7 4 5 8 2 3 0 1 9

  
   1 2 3 4 5 6 14             8 9 10 11 12 13 14 15

   1 2 6    3 4 5 14 

   1 2   6    3 5   4 14       8 9 10 14   11 12 13 15

                               8 10  9 14   11 13   12 15

   Solution: 1 2 6 3 5 4 14 8 10 9 14 11 13 12 15


3. People did this correctly ;-)

4.1 Theta(n lg n) by the Master Theorem 
4.2 Thata (n lg^2 n) by repeated substituition.  Note: the Master theorem
    does NOT apply.
4.3 Theta(n!)
4.4 Theta(n^2)
4.5 Theta(n^3)  (by substition method)