A) Given the equation 3x² − 4x + 2 = 0, with roots alpha and beta.

$$\alpha + \beta = -\frac{-4}{3} = \frac{4}{3}$$ $$\alpha\beta = \frac{2}{3}$$

We know that $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$.
Also, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Substitute this into the identity:

$$\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta]$$ $$\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta]$$

$$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right]$$ $$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - 2\right]$$ $$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[\frac{16}{9} - \frac{18}{9}\right]$$ $$\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left[-\frac{2}{9}\right]$$ $$\alpha^3 + \beta^3 = -\frac{8}{27}$$

This shows that $\boxed{\alpha^3 + \beta^3 = -\frac{8}{27}}$.

$$p+q = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2}$$

First, find $\alpha^2 + \beta^2$:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

Using values from part a): $\alpha + \beta = \frac{4}{3}$ and $\alpha\beta = \frac{2}{3}$.

$$\alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2\left(\frac{2}{3}\right) = \frac{16}{9} - \frac{4}{3} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9}$$

Now substitute this into the sum of new roots:

$$p+q = \frac{\frac{4}{9}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{4}{9}}{\frac{4}{9}} = 1$$

$$pq = \left(\frac{1}{\alpha^2}\right)\left(\frac{1}{\beta^2}\right) = \frac{1}{(\alpha\beta)^2}$$ $$pq = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}$$

$$x^2 - (p+q)x + pq = 0$$ $$x^2 - (1)x + \frac{9}{4} = 0$$

To obtain integral coefficients, multiply the entire equation by 4:

$$4x^2 - 4x + 9 = 0$$

The equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is $\boxed{4x^2 - 4x + 9 = 0}$.