(a) We start with the given equation 3μ = 7σ² and the probability P(X > 2μ) = 0.1016. Standardizing, we have:

\(z = \frac{2μ - μ}{σ} = \frac{μ}{σ}\)

\(From the standard normal distribution table, P(Z > 1.272) = 0.1016, so:\)

\(\frac{7σ}{3} = 1.272\)

Solving for σ, we get:

\(σ = 0.545\)

Substituting back to find μ:

\(μ = \frac{7σ^2}{3} = 0.693\)

(b) Given Y ~ N(33, 21), we need to find a such that P(33 − a < Y < 33 + a) = 0.5. This implies:

\(P(X < a + 33) = 0.75\)

\(From the standard normal distribution table, z = 0.674 for 0.75 probability. Standardizing:\)

\(\frac{a + 33 - 33}{\sqrt{21}} = 0.674\)

Solving for a gives:

\(a = 3.09\)