(i) The probability of taking a blue pen from the left pocket is \(\frac{1}{4}\). After moving this pen to the right pocket, the probability of taking a blue pen from the right pocket is \(\frac{2}{5}\). Therefore, the probability is \(\frac{1}{4} \times \frac{2}{5} = \frac{1}{10}\).

(ii) To have 1 blue pen in the left pocket after operation T, either a red pen is moved from left to right and a red pen is moved back, or a blue pen is moved from left to right and a blue pen is moved back. The probability of moving a red pen from left to right is \(\frac{3}{4}\) and then moving a red pen back is \(\frac{4}{5}\). The probability of moving a blue pen from left to right and then moving a blue pen back is \(\frac{1}{10}\). Thus, \(P(X = 1) = \frac{3}{4} \times \frac{4}{5} + \frac{1}{10} = \frac{14}{20} = \frac{7}{10}\).

(iii) Given that a blue pen is taken from the right pocket, the probability that a blue pen was initially taken from the left pocket is \(\frac{P(B \cap B)}{P(B)} = \frac{1/10}{3/4 \times 1/5 + 1/4 \times 2/5} = \frac{1/10}{1/5} = \frac{2}{5}\).