(a) Let \(y = x^4\). Then the equation becomes \(y + 2y^{3/4} - 1 = 0\). Substituting \(y = (1-y)^4\), we expand using the binomial theorem: \(16y^3 = 1 - 4y + 6y^2 - 4y^3 + y^4\). Rearranging gives \(y^4 - 20y^3 + 6y^2 - 4y + 1 = 0\). The sum \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 20\).

(b) Using \(\alpha + \beta + \gamma + \delta = -2\), multiply the original equation by \(x\) to get \(x^5 + 2x^4 - x = 0\). Thus, \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5 = -2(20) + (-2) = -42\).

(c) Using the formula for the sum of squares, \(\alpha^8 + \beta^8 + \gamma^8 + \delta^8 = 20^2 - 2(6) = 388\).