a) I am satisified, by brute force search

b)
G=[1 0 1 0 0 1;
   0 1 1 0 1 0;
   1 1 0 1 0 0];

c) Verified in Octave. H, as given, doesn't take a transpose. 

d) [1 0 0]
Should be 011->101110

I can correct arbitrary one-bit errors. 

e) We compute the syndrome. The syndrome will be the ith line of the H
matrix. We flip the ith bit in the code vector, and we are left with a
valid code vector. 

If we were slightly more clever about our code, we could have made H
be [0 0 1; 0 1 0; 0 1 1; 1 0 0; 1 0 1 ; 1 1 0 ; 1 1 1], so we would
skip one table lookup.

If there is more than one erronious bit, we are SOL. 

f) Our error rate was 1-.9^6-.9^5*.1*6=11 percent

g) See waterfall.gif.