To solve this problem, we need to calculate the probability distribution for the number of heads obtained when the three coins are tossed.

Let the random variable X represent the number of heads obtained. The possible values of X are 0, 1, 2, and 3.

1. Probability of 0 heads, P(0):
\(P(0) = (1 - 0.4) \times (1 - 0.75) \times (1 - 0.5) = 0.6 \times 0.25 \times 0.5 = 0.075\)

2. Probability of 1 head, P(1):
\(P(1) = 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 + 0.6 \times 0.25 \times 0.5 = 0.35\)

3. Probability of 2 heads, P(2):
\(P(2) = 0.4 \times 0.75 \times 0.5 + 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 = 0.425\)

4. Probability of 3 heads, P(3):
\(P(3) = 0.4 \times 0.75 \times 0.5 = 0.15\)

The probability distribution table is:

To calculate the mean, E(X):
\(E(X) = 0 \times 0.075 + 1 \times 0.35 + 2 \times 0.425 + 3 \times 0.15 = 1.65\)

To calculate the variance, Var(X):
\(Var(X) = (0^2 \times 0.075) + (1^2 \times 0.35) + (2^2 \times 0.425) + (3^2 \times 0.15) - (1.65)^2 = 0.678\)