(i) To find P(X < 2μ), we standardize the variable:

\(P(X < 2μ) = P\left( z < \frac{2μ - μ}{σ} \right) = P\left( z < \frac{μ}{σ} \right).\)

Given 5σ = 3μ, we have \(\frac{μ}{σ} = \frac{5}{3}\).

\(Thus, P(X < 2μ) = P\left( z < \frac{5}{3} \right) = 0.952.\)

(ii) We are given P(X < \(\frac{1}{3}μ\)) = 0.8524. Standardizing gives:

\(P\left( z < \frac{\frac{1}{3}μ - μ}{σ} \right) = P\left( z < \frac{-2μ}{3σ} \right).\)

From the standard normal distribution table, \(\frac{-2μ}{3σ} = -1.047\).

Solving for μ, we get \(\frac{-2μ}{3σ} = -1.047\) which gives \(μ = -1.57σ\).