Use prime factorization method to determine the LCM and HCF of 60 and 80.

The common prime factors are 2 and 5.
For 2, the lowest power is $2^2$. For 5, the lowest power is $5^1$.

$$\text{HCF}(60, 80) = 2^2 \times 5^1 = 4 \times 5 = 20$$

Take all prime factors from both numbers with their highest powers.
For 2, the highest power is $2^4$. For 3, the highest power is $3^1$. For 5, the highest power is $5^1$.

$$\text{LCM}(60, 80) = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 48 \times 5 = 240$$

The HCF is $\boxed{\mathbf{20}}$ and the LCM is $\boxed{\mathbf{240}}$.

Question 3.1.1
Calculate the following to the simplest form: $-15y - (-5y)$.

$$(-15 + 5)y$$ $$-10y$$

The simplest form is $\boxed{\mathbf{-10y}}$.

Question 3.1.2
Calculate $450 \text{ mm}$ of $9 \text{ km}$.

$$1 \text{ km} = 1000 \text{ m}$$ $$1 \text{ m} = 1000 \text{ mm}$$

So, $1 \text{ km} = 1000 \times 1000 \text{ mm} = 1,000,000 \text{ mm}$.

$$9 \text{ km} = 9 \times 1,000,000 \text{ mm} = 9,000,000 \text{ mm}$$

$$\frac{45}{900000} = \frac{1}{20000}$$

The value is $\boxed{\mathbf{\frac{1}{20000}}}$.

Question 3.2.1
Use the distributive property to calculate the following: $8 \times [(-9) + (-7)]$.

$$-72 - 56$$ $$-128$$

The calculated value is $\boxed{\mathbf{-128}}$.

Question 3.3.1
Calculate the following integers: $23nm + (-6nm) - (+20nm) + 3nm$.

$$(23 - 6 - 20 + 3)nm$$ $$(17 - 20 + 3)nm$$ $$(-3 + 3)nm$$ $$0nm$$ $$0$$

The calculated value is $\boxed{\mathbf{0}}$.