Let X be the random variable representing the number of green sweets taken. The possible values of X are 0, 1, and 2.

Calculate the probabilities:

\(P(X = 0) = P(B, B) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}\)

\(P(X = 1) = P(G, B) + P(B, G) = \frac{2}{7} \times \frac{5}{6} + \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}\)

\(P(X = 2) = P(G, G) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21}\)

Calculate the expected value:

\(E(X) = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21} = \frac{4}{7}\)

Calculate the variance:

\(\text{Var}(X) = 0^2 \times \frac{10}{21} + 1^2 \times \frac{10}{21} + 2^2 \times \frac{1}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\)