(i) To find the term in \(x^2\) in the expansion of \((1 - \frac{3}{2}x)^6\), use the binomial theorem:

\(^6C_2 \left( -\frac{3}{2}x \right)^2 = 15 \times \left( \frac{9}{4}x^2 \right) = 135x^2.\)

For the term in \(x^3\):

\(^6C_3 \left( -\frac{3}{2}x \right)^3 = 20 \times \left( -\frac{27}{8}x^3 \right) = -540x^3.\)

(ii) For the expansion of \((k + 2x)(1 - \frac{3}{2}x)^6\), the term in \(x^3\) is:

\(2 \times 135x^2 + k \times (-540x^3) = 0.\)

\(\frac{270x^3}{4} - \frac{135kx^3}{2} = 0.\)

Solving for \(k\):

\(270 - 270k = 0\)

\(k = 1.\)