(i) The total number of balls in box B after transferring one ball from box A is \(5 + x + 1 = x + 6\). Therefore, the probability of choosing a yellow ball from box B is \(\frac{x}{x+6}\).

(ii) The tree diagram is completed as follows:
From box A to box B:
- White to White: \(\frac{6}{x+6}\)
- White to Yellow: \(\frac{x}{x+6}\)
- Yellow to White: \(\frac{5}{x+6}\)
- Yellow to Yellow: \(\frac{x+1}{x+6}\).

(iii) Given that the probability of choosing a white ball from box B after a white ball from box A is \(\frac{1}{3}\), we have:
\(\frac{6}{x+6} = \frac{1}{3}\)
Solving for \(x\):
\(6 = \frac{x+6}{3}\)
\(18 = x + 6\)
\(x = 12\).

(iv) To find the conditional probability that the ball chosen from box A was yellow given that the ball chosen from box B is yellow, we use:
\(P(A_Y | B_Y) = \frac{P(A_Y \cap B_Y)}{P(B_Y)}\)
\(P(B_Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18} = \frac{61}{90}\)
\(P(A_Y \cap B_Y) = \frac{2}{10} \times \frac{13}{18} = \frac{13}{90}\)
\(P(A_Y | B_Y) = \frac{13/90}{61/90} = \frac{13}{61}\).