Brief solutions:

For Q.1 you should have stated that you will decide according to
the max of P(C_i|X), and with Bayes rule you are choosing the max of 
P(X|C_i), since P(x) doesn't depend on (i) and P(C_i) are equal.
By independence, you transform this to a product of P(x_j|C_i) and then
take the log, so now your discriminant function of for class (i) is the
sum of P(x_j|C_i).
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You can write  x_j * p  + (1-x_j)*(1-p) for each term of function_1 and
similarly set up function_2.  Subtract f_2 from f_1 and in 2-3 lines
you're done, showing that if p>0.5 you get a positive answer.
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To rephrase, each discriminant function, f_1 or f_2, is a summation of  j
terms that are either  p  or (1-p).  In f_1 you have  p  for every
observed 1, and (1-p) for every observed 0.  In f_2 the opposite holds. 
Suppose that you have k more observed 1's than 0's.  Then when you
subtract f_2 from f_1, you are left with   k*p - k(1-p)  = k(2p-1).
k is positive and (2p-1) is positive when p>0.5, so f2-f1 is positive.
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For Q.2 there is a counterexample.
You may check the answer to a similar question (involving Gabriel
graph editing instead of RNG editing).  The answer is almost the same.
(question 3 of 2002 exam)
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Q.3 was straight from homework 5.  I believe its the second half of
the third question.