(a) From the roots \(\alpha, \alpha, \gamma\), we have:

\(2\alpha + \gamma = -b\)

\(\alpha^2 + 2\alpha\gamma = 0\)

Solving, \(\alpha = -2\gamma\) leading to \(-4\gamma + \gamma = -b\), so \(\gamma = \frac{1}{3}b\) and \(\alpha = -\frac{2}{3}b\).

\(\alpha^2\gamma = -d\) leading to \(\frac{4}{27}b^3 = -d\) leading to \(4b^3 + 27d = 0\).

(b) Given \(2\alpha^2 + \gamma^2 = 3b\), substitute for \(\alpha\) and \(\gamma\):

\(3b = b^2\) leading to \(b = 3\).

Substitute back to find \(d\):

\(d = -4\).