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

\(y^2 - y + 2y^{\frac{1}{2}} + 5 = 0\).

Rearrange and square both sides to eliminate the square root:

\((2y^{\frac{1}{2}})^2 = (-y^2 + y - 5)^2\).

Expand and simplify to obtain:

\(y^4 - 2y^3 + 11y^2 - 14y + 25 = 0\).

The value of \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\) is 2.

(b) Using the relation \(\alpha^2 \beta^2 \gamma^2 \delta^2 = 25\), we have:

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

Substitute the known values to get \(\frac{14}{25}\).

(c) Using the identity:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)^2 - 2(\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \alpha^2 \delta^2 + \beta^2 \gamma^2 + \beta^2 \delta^2 + \gamma^2 \delta^2)\).

Substitute the known values to get \(2^2 - 2(11) = -18\).