Find the value of the following limits: 5. lim (x->2) [1/(x(x-2)2) - 1/(x2-3x+2)]; 6. lim (x->1) [1/(1-x) - 3/(1-x3)]

The quadratic $x^2-3x+2$ factors as $(x-1)(x-2)$.  
The expression becomes:
$$
\lim_{x \to 2} \left[ \frac{1}{x(x-2)^2} - \frac{1}{(x-1)(x-2)} \right]
$$

The common denominator is $x(x-1)(x-2)^2$.
$$
= \lim_{x \to 2} \left[ \frac{(x-1)}{x(x-1)(x-2)^2} - \frac{x(x-2)}{x(x-1)(x-2)^2} \right]
$$
$$
= \lim_{x \to 2} \frac{(x-1) - x(x-2)}{x(x-1)(x-2)^2}
$$

$$
= \lim_{x \to 2} \frac{x-1 - x^2 + 2x}{x(x-1)(x-2)^2}
$$
$$
= \lim_{x \to 2} \frac{-x^2 + 3x - 1}{x(x-1)(x-2)^2}
$$

Substitute $x=2$ into the numerator: $-(2)^2 + 3(2) - 1 = -4 + 6 - 1 = 1$.  
Substitute $x=2$ into the denominator: $2(2-1)(2-2)^2 = 2(1)(0)^2 = 0$.  
Since the numerator approaches 1 and the denominator approaches 0 from the positive side (because $(x-2)^2$ is always positive, and $x(x-1)$ is positive near $x=2$), the limit is $\frac{1}{0^+}$.  
The value of the limit is $\boxed{\infty}$.

The difference of cubes formula is $a^3-b^3 = (a-b)(a^2+ab+b^2)$. So, $1-x^3 = (1-x)(1+x+x^2)$.  
The expression becomes:
$$
\lim_{x \to 1} \left( \frac{1}{1-x} - \frac{3}{(1-x)(1+x+x^2)} \right)
$$

The common denominator is $(1-x)(1+x+x^2)$.
$$
= \lim_{x \to 1} \frac{(1+x+x^2) - 3}{(1-x)(1+x+x^2)}
$$

$$
= \lim_{x \to 1} \frac{x^2 + x - 2}{(1-x)(1+x+x^2)}
$$

The numerator $x^2+x-2$ factors as $(x+2)(x-1)$.  
Also, note that $(1-x) = -(x-1)$.
$$
= \lim_{x \to 1} \frac{(x+2)(x-1)}{-(x-1)(1+x+x^2)}
$$
Cancel out the common factor $(x-1)$ (since $x \ne 1$ as $x \to 1$):
$$
= \lim_{x \to 1} \frac{x+2}{-(1+x+x^2)}
$$

$$
= \frac{1+2}{-(1+1+1^2)} = \frac{3}{-(1+1+1)} = \frac{3}{-3} = -1
$$
The value of the limit is $\boxed{-1}$.

Substitute $x=\sqrt{2}$ into the numerator: $(\sqrt{2})^4 - 4 = 4 - 4 = 0$.  
Substitute $x=\sqrt{2}$ into the denominator: $(\sqrt{2})^2 + 3\sqrt{2}(\sqrt{2}) - 8 = 2 + 3(2) - 8 = 2 + 6 - 8 = 0$.  
This is an indeterminate form $\frac{0}{0}$, so we need to factor.

The numerator is a difference of squares: $x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2)$.  
Further factor $x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2})$.  
So, $x^4 - 4 = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)$.

Since substituting $x=\sqrt{2}$ into the denominator yields 0, $(x-\sqrt{2})$ must be a factor.  
We can find the other factor by polynomial division or by inspection:  
$x^2 + 3\sqrt{2}x - 8 = (x - \sqrt{2})(x + 4\sqrt{2})$.

$$
= \lim_{x \to \sqrt{2}} \frac{(x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)}{(x - \sqrt{2})(x + 4\sqrt{2})}
$$

Cancel out $(x - \sqrt{2})$ (since $x \ne \sqrt{2}$ as $x \to \sqrt{2}$):
$$
= \lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2})(x^2 + 2)}{x + 4\sqrt{2}}
$$
Now substitute $x=\sqrt{2}$:
$$
= \frac{(\sqrt{2} + \sqrt{2})((\sqrt{2})^2 + 2)}{\sqrt{2} + 4\sqrt{2}}
$$
$$
= \frac{(2\sqrt{2})(2 + 2)}{5\sqrt{2}}
$$
$$
= \frac{(2\sqrt{2})(4)}{5\sqrt{2}}
$$
$$
= \frac{8\sqrt{2}}{5\sqrt{2}} = \frac{8}{5}
$$
The value of the limit is $\boxed{\frac{8}{5}}$.