Given universal set mu = Z, P = \..., -4, -2, 0, 2, 4, ... Q = : -7 < x < 7 and R = : x > 0 find P R, P' Q, and n(Q R).

This means finding the elements that are both even and positive.

$$P \cap R = \{x : x \text{ is an even integer and } x > 0\}$$ $$P \cap R = \{2, 4, 6, 8, ...\}$$

The intersection is $\boxed{\{2, 4, 6, 8, ...\}}$.

b) $P' \cap Q$

Since $P$ is the set of all even integers, $P'$ is the set of all odd integers.

$$P' = \{..., -5, -3, -1, 1, 3, 5, ...\}$$

This means finding the elements that are both odd and within the range of $Q$.

$$P' \cap Q = \{-5, -3, -1, 1, 3, 5\}$$

The intersection is $\boxed{\{-5, -3, -1, 1, 3, 5\}}$.

c) $n(Q \cap R)$

$$Q \cap R = \{x : -7 < x < 7 \text{ and } x > 0, x \in \mathbb{Z}\}$$ $$Q \cap R = \{1, 2, 3, 4, 5, 6\}$$

$$n(Q \cap R) = 6$$

The number of elements is $\boxed{6}$.

:
A binary operation $*$ is defined on the set of real numbers, $\mathbb{R}$, by $p * q = \frac{3p + 2q}{2p + 3q}$.

a) Evaluate $(\sqrt{3}) * (\sqrt{2})$.

Numerator:

$$(3\sqrt{3} + 2\sqrt{2})(2\sqrt{3} - 3\sqrt{2}) = (3\sqrt{3})(2\sqrt{3}) - (3\sqrt{3})(3\sqrt{2}) + (2\sqrt{2})(2\sqrt{3}) - (2\sqrt{2})(3\sqrt{2})$$ $$= (6 \times 3) - 9\sqrt{6} + 4\sqrt{6} - (6 \times 2)$$ $$= 18 - 9\sqrt{6} + 4\sqrt{6} - 12$$ $$= 6 - 5\sqrt{6}$$

Denominator:

$$(2\sqrt{3} + 3\sqrt{2})(2\sqrt{3} - 3\sqrt{2}) = (2\sqrt{3})^2 - (3\sqrt{2})^2$$ $$= (4 \times 3) - (9 \times 2)$$ $$= 12 - 18$$ $$= -6$$

$$(\sqrt{3}) * (\sqrt{2}) = \frac{6 - 5\sqrt{6}}{-6}$$ $$(\sqrt{3}) * (\sqrt{2}) = \frac{-(6 - 5\sqrt{6})}{6}$$ $$(\sqrt{3}) * (\sqrt{2}) = \frac{-6 + 5\sqrt{6}}{6}$$

The evaluation is $\boxed{\frac{5\sqrt{6} - 6}{6}}$.