Out of 42 points

1: (6pt)
  m2l = c2r xor c1r xor m1l =  832B4AF7

2: (2pt each)
  A) repeated outputs
  B) right rotate of bits
  C) b0 | b1 | b2 -> ~b2 | .. | b0 xor b1 xor b2
     (also accepted answers based on differential analysis)

3: (3pt each)
  A) x^{(p-1)(q-1) - 1} 
  B) 22
  C) 105^235 mod 391 = 41
  D) ed = k(p-1)(q-1) + 1
        = k(n/q -1)(q-1) + 1
        = k(n - q -n/q -1) + 1 
      k <= 3 (for e = 3)
     solve for q by trying each k

4: (3pt each)
  A) (2,3)
  B) 6
  C) 10 is not a generator for 11
     In fact, in general (p-1)^k mod p will
     only generate 1 and -1 since p-1 = -1 mod p in
     modular arithmetic.

5: (3pt and 6pt)
  A) 39, 58, 53, 37, 60, 58, 53, ...
      1,  0,  1,  1,  0,  0,  1, ...
  B) (p-1)(q-1)/4 got full credit if explained.
     In particular since (p-1) and (q-1) are each divisible by 2
     2^i mod (p-1)(q-1) will generate at most (p-1)(q-1)/4 + 2 
     values.
     (p-1)(q-1) and (p-1)(q-1)/2 got partial credit.