(a) To find the probability that a randomly chosen male leopard weighs between 46 and 62 kg, we standardize the values using the formula:

\(P(46 < X < 62) = P\left( \frac{46 - 55}{6} < Z < \frac{62 - 55}{6} \right)\)

This simplifies to:

\(P(-1.5 < Z < 1.1667)\)

Using the standard normal distribution table, we find:

\(\Phi(1.1667) - \Phi(-1.5) = 0.8784 + (0.9332 - 1) = 0.812\)

(b) Given that 25% of female leopards weigh less than 36 kg, we find the z-value corresponding to 0.25, which is approximately -0.674. Using the standardization formula:

\(\frac{36 - 42}{\sigma} = -0.674\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{36 - 42}{-0.674} = 8.9\)

(c) To find the probability that both leopards weigh less than 46 kg, we calculate:

For male leopards: \(P(X < 46) = 1 - \Phi\left(\frac{46 - 55}{6}\right) = 0.0668\)

For female leopards: \(P(Y < 46) = \Phi\left(\frac{46 - 42}{8.9}\right) = 0.6732\)

The probability that both events occur is:

\(P(\text{both}) = 0.0668 \times 0.6732 = 0.0450\)