To solve part (i), start with the given equation:

\(y^3 = Ae^{2x}\)

Take the natural logarithm of both sides:

\(\ln(y^3) = \ln(Ae^{2x})\)

Using logarithm properties, this becomes:

\(3\ln(y) = \ln(A) + 2x\)

This is in the form \(y = mx + c\), where the gradient \(m\) is \(\frac{2}{3}\).

For part (ii), substitute \(x = 0\) and \(\ln(y) = 0.5\) into the equation:

\(3(0.5) = \ln(A) + 2(0)\)

\(1.5 = \ln(A)\)

Exponentiate both sides to solve for \(A\):

\(A = e^{1.5}\)

Calculate \(A\) to two decimal places:

\(A \approx 4.48\)