The problem asks us to find the expressions from (a) to (e) whose values (product or quotient) are greater than $\frac{2}{3}$.

1. Problem Description

The problem asks us to find the expressions from (a) to (e) whose values (product or quotient) are greater than $\frac{2}{3}$.

2. Solution Steps

First, we evaluate each expression:

(a) $\frac{2}{3} \times \frac{1}{8} = \frac{2 \times 1}{3 \times 8} = \frac{2}{24} = \frac{1}{12}$

Since $\frac{1}{12} < \frac{2}{3}$, this expression is not a solution.

(i) $\frac{2}{3} \times \frac{9}{7} = \frac{2 \times 9}{3 \times 7} = \frac{18}{21} = \frac{6}{7}$

We compare $\frac{6}{7}$ and $\frac{2}{3}$ by finding a common denominator:

$\frac{6}{7} = \frac{6 \times 3}{7 \times 3} = \frac{18}{21}$

$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

Since $\frac{18}{21} > \frac{14}{21}$, i.e. $\frac{6}{7} > \frac{2}{3}$, this expression is a solution.

(u) $\frac{2}{3} \div \frac{3}{5} = \frac{2}{3} \times \frac{5}{3} = \frac{2 \times 5}{3 \times 3} = \frac{10}{9}$

Since $\frac{10}{9} > \frac{2}{3}$, this expression is a solution.

We compare $\frac{10}{9}$ and $\frac{2}{3}$ by finding a common denominator:

$\frac{10}{9}$

$\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$

Since $\frac{10}{9} > \frac{6}{9}$, i.e. $\frac{10}{9} > \frac{2}{3}$, this expression is a solution.

(e) $\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}$

We compare $\frac{8}{15}$ and $\frac{2}{3}$ by finding a common denominator:

$\frac{8}{15}$

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

Since $\frac{8}{15} < \frac{10}{15}$, i.e. $\frac{8}{15} < \frac{2}{3}$, this expression is not a solution.

The expressions (i) and (u) are greater than $\frac{2}{3}$.

3. Final Answer

i, u