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

For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha \beta = \frac{c}{a}$.
Here, $a=3$, $b=-4$, $c=2$.

The identity is $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)$.

This shows the required result.

1(i)b) Find the equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$

Let the new roots be $r_1 = \frac{1}{\alpha^2}$ and $r_2 = \frac{1}{\beta^2}$.
The general quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.

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

Substitute the values from part (a):

Now, calculate $S'$:

Substitute the value of $\alpha \beta$:

The equation is $x^2 - S'x + P' = 0$.

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

The equation with integral coefficients is \boxed{\text{4x^2 - 4x + 9 = 0}}.

1(ii) Find the set of real values of $x$ for which $\left|\frac{x+3}{x-1}\right| = \frac{x+3}{x-1}$

The property $|A| = A$ holds true if and only if $A \ge 0$.
Therefore, we need to find the values of $x$ for which $\frac{x+3}{x-1} \ge 0$.
Also, the denominator cannot be zero, so $x-1 \ne 0 \implies x \ne 1$.

For the fraction to be non-negative, both numerator and denominator must have the same sign (or the numerator is zero).

Case 1: Numerator is non-negative and denominator is positive.
$x+3 \ge 0 \implies x \ge -3$
$x-1 > 0 \implies x > 1$
The intersection of these conditions is $x > 1$.

Case 2: Numerator is non-positive and denominator is negative.
$x+3 \le 0 \implies x \le -3$
$x-1 < 0 \implies x < 1$
The intersection of these conditions is $x \le -3$.

The set of real values of $x$ is $x \le -3$ or $x > 1$.
In interval notation, this is $(-\infty, -3] \cup (1, \infty)$.

The set of real values of $x$ is $\boxed{x \le -3 \text{ or } x > 1}$.