Use the logarithm property ((A)/(B)) = A − B.

Here are the solutions to the problems.

1. (a) Given that $\log m = x$ and $\log n = 2x - 1$, find an expression for $\log \left(\frac{mn^2}{100}\right)$.

Step 1: Use the logarithm property $\log\left(\frac{A}{B}\right) = \log A - \log B$.

$$\log \left(\frac{mn^2}{100}\right) = \log(mn^2) - \log(100)$$

Step 2: Use the logarithm properties $\log(AB) = \log A + \log B$ and $\log(A^k) = k \log A$.

$$\log(mn^2) - \log(100) = \log m + \log(n^2) - \log(100) = \log m + 2 \log n - \log(100)$$

Step 3: Substitute the given values for $\log m$ and $\log n$, and evaluate $\log(100)$ (assuming base 10 logarithm).

$$x + 2(2x - 1) - 2$$

Step 4: Simplify the expression.

$$x + 4x - 2 - 2 = 5x - 4$$

The expression is $\boxed{\mathbf{5x - 4}}$.

1. (b) Simplify: $(1 - \sqrt{5})\left(\frac{1}{5} + \sqrt{5}\right)$ leaving your answer in the form $m + n\sqrt{5}$, where $m$ and $n$ are rational numbers.

Step 1: Expand the product using the distributive property (FOIL).

$$(1 - \sqrt{5})\left(\frac{1}{5} + \sqrt{5}\right) = 1 \cdot \frac{1}{5} + 1 \cdot \sqrt{5} - \sqrt{5} \cdot \frac{1}{5} - \sqrt{5} \cdot \sqrt{5}$$

Step 2: Simplify each term.

$$= \frac{1}{5} + \sqrt{5} - \frac{\sqrt{5}}{5} - 5$$

Step 3: Group the rational terms and the terms containing $\sqrt{5}$.

$$= \left(\frac{1}{5} - 5\right) + \left(\sqrt{5} - \frac{\sqrt{5}}{5}\right)$$

Step 4: Perform the arithmetic for each group.

$$= \left(\frac{1}{5} - \frac{25}{5}\right) + \left(\frac{5\sqrt{5}}{5} - \frac{\sqrt{5}}{5}\right)$$ $$= -\frac{24}{5} + \frac{4\sqrt{5}}{5}$$

Step 5: Express the answer in the form $m + n\sqrt{5}$.

$$= -\frac{24}{5} + \frac{4}{5}\sqrt{5}$$

The simplified expression is $\boxed{\mathbf{-\frac{24}{5} + \frac{4}{5}\sqrt{5}}}$.

2. (a) Construct a multiplication $\otimes$ table for modulo 9 on the set $\{2, 4, 6, 8\}$.

Step 1: Create a table with the given elements as headers for rows and columns.

Step 2: Calculate each product $a \times b \pmod{9}$ for $a, b \in \{2, 4, 6, 8\}$ and fill the table.
For example, $2 \otimes 2 = (2 \times 2) \pmod{9} = 4 \pmod{9} = 4$.
$2 \otimes 6 = (2 \times 6) \pmod{9} = 12 \pmod{9} = 3$.

The multiplication table is:

\begin{array}{|c|c|c|c|c|} \hline \otimes & 2 & 4 & 6 & 8 \\ \hline 2 & 4 & 8 & 3 & 7 \\ \hline 4 & 8 & 7 & 6 & 5 \\ \hline 6 & 3 & 6 & 0 & 3 \\ \hline 8 & 7 & 5 &