Simplify sqrt(8).

$$(2\sqrt{2}-2)(4+\sqrt{2}) = 2\sqrt{2}(4) + 2\sqrt{2}(\sqrt{2}) - 2(4) - 2(\sqrt{2})$$ $$= 8\sqrt{2} + 2(2) - 8 - 2\sqrt{2}$$ $$= 8\sqrt{2} + 4 - 8 - 2\sqrt{2}$$ $$= (8\sqrt{2} - 2\sqrt{2}) + (4 - 8)$$ $$= 6\sqrt{2} - 4$$

$$(4-\sqrt{2})(4+\sqrt{2}) = 4^2 - (\sqrt{2})^2$$ $$= 16 - 2$$ $$= 14$$

$$\frac{2(3\sqrt{2}-2)}{2(7)}$$ $$= \frac{3\sqrt{2}-2}{7}$$

The simplified expression is $\boxed{\frac{3\sqrt{2}-2}{7}}$.

b) To express the given expression as a single fraction, we first factor the denominator of the first term.

We look for two numbers that multiply to -2 and add to -1. These numbers are -2 and 1.

$$y^2-y-2 = (y-2)(y+1)$$

To get this common denominator for the second term, multiply its numerator and denominator by $(y-2)$.

$$\frac{1}{y+1} = \frac{1}{y+1} \times \frac{y-2}{y-2} = \frac{y-2}{(y-2)(y+1)}$$

$$\frac{1}{(y-2)(y+1)} - \frac{y-2}{(y-2)(y+1)}$$ $$= \frac{1 - (y-2)}{(y-2)(y+1)}$$

$$\frac{3-y}{(y-2)(y+1)}$$

The expression as a single fraction is $\boxed{\frac{3-y}{(y-2)(y+1)}}$.

a) Let the first term of the Arithmetic Progression (AP) be $a$ and the common difference be $d$.
The first three terms are $a$, $a+d$, and $a+2d$.

The average of the first three terms is 56.

$$\frac{a + (a+d) + (a+2d)}{3} = 56$$ $$\frac{3a + 3d}{3} = 56$$ $$a + d = 56 \quad (*)$$

The sum of the first two terms is 96.

$$a + (a+d) = 96$$ $$2a + d = 96 \quad (**)$$

Subtract equation $(*)$ from equation $(**)$:

$$(2a + d) - (a + d) = 96 - 56$$ $$a = 40$$

$$40 + d = 56$$ $$d = 56 - 40$$ $$d = 16$$

The first term is $\boxed{40}$ and the common difference is $\boxed{16}$.