Given that the lengths of fish are normally distributed, we have:

\(P(X < 12) = \frac{42}{400} = 0.105\)

\(P(X > 19) = \frac{58}{400} = 0.145\)

Using the standard normal distribution table, we find the corresponding z-values:

\(\frac{12 - \mu}{\sigma} = -1.253\)

\(\frac{19 - \mu}{\sigma} = 1.058\)

We now have two equations:

\(12 - \mu = -1.253\sigma\)

\(19 - \mu = 1.058\sigma\)

Subtract the first equation from the second:

\(19 - 12 = 1.058\sigma + 1.253\sigma\)

\(7 = 2.311\sigma\)

Solving for \(\sigma\):

\(\sigma = \frac{7}{2.311} \approx 3.03\)

Substitute \(\sigma\) back into one of the equations to find \(\mu\):

\(12 - \mu = -1.253 \times 3.03\)

\(12 - \mu = -3.797\)

\(\mu = 12 + 3.797 \approx 15.8\)

Thus, the estimates for the mean and standard deviation are \(\mu = 15.8\) and \(\sigma = 3.03\).