#### Answer

Simplified expression: $\frac{2(a-1)}{3(a+1)}$
Excluded values: $-1$ and $1$.

#### Work Step by Step

$\frac{2a^2-4a+2}{3a^2-3}$
In a rational expression, excluded values are any real numbers that make the denominator equal to zero. In this case, any solutions to the equation $3a^2-3=0$ would be excluded values.
$3a^2-3=0$
$3(a^2-1)=0$
$3(a-1)(a+1)=0$
$a-1=0$ or $a+1=0$
$a=1$ or $a=-1$
Therefore, -1 and 1 are two excluded values.
Now, we can find the simplified expression:
$\frac{2a^2-4a+2}{3a^2-3}$
$=\frac{2(a^2-2a+1)}{3(a-1)(a+1)}$
$=\frac{2(a-1)^2}{3(a-1)(a+1)}$
$=\frac{2(a-1)}{3(a+1)}$