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Let's solve each part of this question by finding the value of \( k \).

### Part (i)

We have:
\[
\left( \frac{2}{3} \right)^3 \times \left( \frac{2}{3} \right)^6 = \left( \frac{4}{9} \right)^{2k - 3}
\]

To simplify, let’s start by applying the property of exponents:
\[
\left( \frac{2}{3} \right)^3 \times \left( \frac{2}{3} \right)^6 = \left( \frac{2}{3} \right)^{3 + 6} = \left( \frac{2}{3} \right)^9
\]

Next, we recognize that \( \frac{4}{9} \) is equivalent to \( \left( \frac{2}{3} \right)^2 \). Thus, we can rewrite the right side as:
\[
\left( \frac{4}{9} \right)^{2k - 3} = \left( \left( \frac{2}{3} \right)^2 \right)^{2k - 3} = \left( \frac{2}{3} \right)^{2(2k - 3)}
\]

Now we have:
\[
\left( \frac{2}{3} \right)^9 = \left( \frac{2}{3} \right)^{2(2k - 3)}
\]

Equating the exponents:
\[
9 = 2(2k - 3)
\]
Expanding and solving for \( k \):
\[
9 = 4k - 6
\]
\[
15 = 4k
\]
\[
k = \frac{15}{4} = 3.75
\]

So, \( k = 3.75 \) for part (i).

---

### Part (ii)

We have:
\[
\left( -\frac{4}{5} \right)^2 \times \left( \frac{4}{5} \right)^5 = \left( \frac{4}{5} \right)^{6k + 1}
\]

First, let’s simplify the left side. Note that:
\[
\left( -\frac{4}{5} \right)^2 = \left( \frac{4}{5} \right)^2
\]

Thus,
\[
\left( \frac{4}{5} \right)^2 \times \left( \frac{4}{5} \right)^5 = \left( \frac{4}{5} \right)^{2 + 5} = \left( \frac{4}{5} \right)^7
\]

Now we have:
\[
\left( \frac{4}{5} \right)^7 = \left( \frac{4}{5} \right)^{6k + 1}
\]

Equating the exponents:
\[
7 = 6k + 1
\]
Solving for \( k \):
\[
6 = 6k
\]
\[
k = 1
\]

So, \( k = 1 \) for part (ii).

---

### Final Answers
- **Part (i):** \( k = 3.75 \)
- **Part (ii):** \( k = 1 \)

Would you like further details on any of the steps? 

Here are some related questions you might explore:
1. How do exponent properties apply when multiplying terms with the same base?
2. Why is it necessary to convert \( \frac{4}{9} \) into \( \left( \frac{2}{3} \right)^2 \)?
3. How would the solution change if the bases were not the same?
4. Can you solve similar problems with different fractions?
5. How do negative exponents affect the calculations in such problems?

**Tip:** Remember, when bases are the same in exponent equations, you can equate the exponents directly to simplify the solution.

Find the value of k in each of the following equations: (i) \(\left(\frac{2}{3}\right)^3 \times \left(\frac{2}{3}\right)^6 = \left(\frac{4}{9}\right)^{2k - 3}\) and (ii) \(\left(-\frac{4}{5}\right)^2 \times \left(\frac{4}{5}\right)^5 = \left(\frac{4}{5}\right)^{6k + 1}\).

Learn how to solve for the variable k in exponential equations with fractional bases, covering key algebra concepts such as the product of powers and power of a power properties. This problem is suitable for students in Grades 9-10.

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