(i) The probability that the first ball is red is \(\frac{3}{8}\). After removing one red ball, the probability that the second ball is red is \(\frac{2}{7}\). Thus, the probability that both balls are red is \(\frac{3}{8} \times \frac{2}{7} = \frac{3}{28}\).

(ii) The probability of choosing a red then a white ball is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). The probability of choosing a white then a red ball is \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Therefore, the probability of different colors is \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(iii) Using conditional probability, \(P(\text{first red} | \text{second red}) = \frac{P(\text{first red and second red})}{P(\text{second red})} = \frac{\frac{3}{28}}{\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}} = \frac{2}{7}\).

(iv) The probability distribution table for \(X\) is:

(v) The expected value \(E(X) = \frac{30}{56} = \frac{15}{28}\). The variance \(\text{Var}(X) = \frac{30}{56} - \left(\frac{15}{28}\right)^2 = \frac{45}{112}\) or 0.402.