(a) By Vieta's formulas, the sum of the products of the roots taken two at a time is given by \(-\frac{b}{a}\) for the equation \(ax^3 + bx^2 + cx + d = 0\). Here, \(a = 2\), \(b = 1\), so \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{1}{2}p\).

(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factored as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). From Vieta's formulas, \(\alpha\beta\gamma = \frac{5}{2}\) and \(\alpha + \beta + \gamma = -\frac{1}{2}\). Therefore, \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \frac{5}{2} \times -\frac{1}{2} = -\frac{5}{4}\).

(c) The roots of the new cubic equation are \(\alpha\beta, \beta\gamma, \alpha\gamma\). The sum of these roots is \(\alpha\beta + \beta\gamma + \alpha\gamma = -\frac{1}{2}p\), the sum of the products of the roots taken two at a time is \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = -\frac{5}{4}\), and the product of the roots is \((\alpha\beta)(\beta\gamma)(\alpha\gamma) = (\alpha\beta\gamma)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}\). The cubic equation is \(4z^3 + 2pz^2 - 5z - 25 = 0\).

(d) Given \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), we use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the known values, \(\frac{1}{3} = \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}p\right)\). Solving for \(p\), \(\frac{1}{3} = \frac{1}{4} + p\), so \(p = \frac{1}{12}\).