(a) Let \(y = x^{-2}\), then \(x = y^{-\frac{1}{2}}\). Substitute into the original equation:

\(y^{-2} + 3y^{-1} + 2y^{-\frac{1}{2}} + 6 = 0\)

Multiply through by \(y^2\) to clear fractions:

\(1 + 3y + 2y^{\frac{3}{2}} + 6y^2 = 0\)

Rearrange to form a quartic equation:

\(36y^4 + 32y^3 + 21y^2 + 6y + 1 = 0\)

The value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\) is \(-\frac{8}{9}\).

(b) Using the relation \(\alpha \beta \gamma \delta = 6\), we find:

\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2} = \frac{\alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 \delta^2}{\alpha^2 \beta^2 \gamma^2 \delta^2}\)

\(\beta^2 \gamma^2 \delta^2 + \alpha^2 \gamma^2 \delta^2 + \alpha^2 \beta^2 \delta^2 + \alpha^2 \beta^2 \gamma^2 = -32\)

(c) Using the formula for the sum of squares:

\(\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} = \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} + \frac{1}{\delta^2}\right)^2 - 2(\alpha^{-2}\beta^{-2} + \alpha^{-2}\gamma^{-2} + \alpha^{-2}\delta^{-2} + \beta^{-2}\gamma^{-2} + \beta^{-2}\delta^{-2} + \gamma^{-2}\delta^{-2})\)

\(= \left(-\frac{8}{9}\right)^2 - 2\left(\frac{21}{36}\right)\)

= \(-\frac{61}{162}\)