(a) Let the probability that a student sings in the choir be denoted by P(C). We know that:

\(P(not C) = 0.58\)

\(Therefore, P(C) = 1 - 0.58 = 0.42\)

\(The probability that a student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that a student does not play in the band and sings in the choir is 0.4 × x.

Thus, the total probability of singing in the choir is:

\(0.18 + 0.4x = 0.42\)

Solving for x:

\(0.4x = 0.42 - 0.18\)

\(0.4x = 0.24\)

\(x = \frac{0.24}{0.4} = 0.6\)

\((b) The probability that one student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that both students play in the band and both sing in the choir is:

\((0.18)^2 = 0.0324\)