(a) Consider the expansion of \((2x^2 + \frac{a}{x})^6\). The general term is given by:

\(\binom{6}{r} (2x^2)^{6-r} \left(\frac{a}{x}\right)^r\)

For \(x^6\), we have:

\(\binom{6}{2} (2x^2)^4 \left(\frac{a}{x}\right)^2 = 15 \times 16x^8 \times \frac{a^2}{x^2} = 15 \times 16a^2 x^6\)

For \(x^3\), we have:

\(\binom{6}{3} (2x^2)^3 \left(\frac{a}{x}\right)^3 = 20 \times 8x^6 \times \frac{a^3}{x^3} = 20 \times 8a^3 x^3\)

Equating the coefficients:

\(15 \times 16a^2 = 20 \times 8a^3\)

\(240a^2 = 160a^3\)

\(a = \frac{3}{2}\)

(b) In the expansion of \((1-x^3)(2x^2 + \frac{a}{x})^6\), the term \(x^6\) comes from the product of \(-x^3\) and the constant term in \((2x^2 + \frac{a}{x})^6\), which is zero. Therefore, the coefficient of \(x^6\) is 0.