(a) We use the binomial distribution with parameters \(n = 12\) and \(p = 0.72\). We need to find \(P(X \leq 9)\), which is \(1 - P(X \geq 10)\).

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\)

\(P(X = 10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(X = 11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(X = 12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Calculating these probabilities:

\(P(X = 10) = 0.19372\)

\(P(X = 11) = 0.09057\)

\(P(X = 12) = 0.01941\)

\(P(X \geq 10) = 0.19372 + 0.09057 + 0.01941 = 0.30304\)

\(P(X \leq 9) = 1 - 0.30304 = 0.696\)

(b) The probability that Moena does not message Pasha for 3 consecutive days and then messages on the 4th day is:

\((0.28)^3 \times 0.72 = 0.0158\)