I) Find the relation connecting R, L and r. ii) Calculate the resistance of the same type of copper wire 750m long and radius 0.25cm.

The resistance $R$ varies directly as its length $L$, so $R \propto L$.
The resistance $R$ varies inversely as the square of its radius $r$, so $R \propto \frac{1}{r^2}$.
Combining these, we get:

$$R \propto \frac{L}{r^2}$$

Given: $R = 30 \text{ ohms}$, $L = 500 \text{ m}$, $r = 0.2 \text{ cm}$.

$$30 = \frac{k \times 500}{(0.2)^2}$$ $$30 = \frac{500k}{0.04}$$ $$30 \times 0.04 = 500k$$ $$1.2 = 500k$$ $$k = \frac{1.2}{500} = \frac{12}{5000} = \frac{3}{1250}$$

The relation connecting $R, L$ and $r$ is:

$$\boxed{R = \frac{3L}{1250r^2}}$$

Given: $L = 750 \text{ m}$, $r = 0.25 \text{ cm}$, and $k = \frac{3}{1250}$.

$$R = \frac{kL}{r^2}$$ $$R = \frac{\frac{3}{1250} \times 750}{(0.25)^2}$$

$$R = \frac{3 \times 750}{1250 \times (0.25)^2}$$ $$R = \frac{2250}{1250 \times 0.0625}$$ $$R = \frac{2250}{78.125}$$

Alternatively, simplify the fraction:

$$R = \frac{3 \times 750}{1250 \times (1/4)^2}$$ $$R = \frac{3 \times 3 \times 250}{5 \times 250 \times (1/16)}$$ $$R = \frac{9}{5 \times (1/16)}$$ $$R = \frac{9}{5/16}$$ $$R = \frac{9 \times 16}{5}$$ $$R = \frac{144}{5}$$ $$R = 28.8$$

The resistance of the wire is:

$$\boxed{28.8 \text{ ohms}}$$

$$\frac{105}{32} \times \frac{320}{21} = \frac{105}{21} \times \frac{320}{32}$$ $$= 5 \times 10$$ $$= 50$$

$$\frac{105}{32} \times \frac{864}{15} = \frac{105}{15} \times \frac{864}{32}$$ $$= 7 \times 27$$ $$= 189$$

$$50 + 189 = 239$$

The simplified expression is:

$$\boxed{239}$$

$$2^3 \times 3^2 \times 3^{-4} \times 2^{4 \times \frac{1}{2}}$$ $$2^3 \times 3^2 \times 3^{-4} \times 2^2$$

$$(2^3 \times 2^2) \times (3^2 \times 3^{-4})$$ $$2^{3+2} \times 3^{2-4}$$ $$2^5 \times 3^{-2}$$

$$32 \times \frac{1}{3^2}$$ $$32 \times \frac{1}{9}$$

$$\frac{32}{9}$$

The simplified expression in its lowest term is:

$$\boxed{\frac{32}{9}}$$