(a) Using Vieta's formulas, we have:

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

\(5 = (-3)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\).

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived equation and given \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27\):

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).