(i) Draw a rough sketch of `y = sin 2x ` and determine the area enclosed by the curve, x-axis and the lines `x = (pi)/(4) and x = (3pi)/(4).`
(ii) Draw the graph of `y = cos 3x, 0 lt x le (pi)/(6)` and find the area between the curve and the axes.

Answer

### Step-by-Step Solution

#### Part (i): Area Enclosed by the Curve \( y = \sin(2x) \)

1. **Determine the Period of the Function**:
The function \( y = \sin(2x) \) has a period given by \( \frac{2\pi}{n} \) where \( n = 2 \). Thus, the period is:
\[ T = \frac{2\pi}{2} = \pi \]

2. **Identify the Points of Interest**:
We need to find the area between the curve and the x-axis from \( x = \frac{\pi}{4} \) to \( x = \frac{3\pi}{4} \).

3. **Sketch the Graph**:
The graph of \( y = \sin(2x) \) oscillates between -1 and 1. At \( x = \frac{\pi}{4} \), \( y = \sin\left(2 \cdot \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \). At \( x = \frac{3\pi}{4} \), \( y = \sin\left(2 \cdot \frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \).

4. **Set Up the Integral**:
The area \( A \) can be calculated using the integral:
\[ A = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} |\sin(2x)| \, dx \]
Since \( \sin(2x) \) is positive from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) and negative from \( \frac{\pi}{2} \) to \( \frac{3\pi}{4} \), we can split the integral:
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2x) \, dx - \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin(2x) \, dx \]

5. **Calculate the First Integral**:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]
Evaluating from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \):
\[ \left[-\frac{1}{2} \cos(2x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\frac{1}{2} \left(\cos(\pi) - \cos\left(\frac{\pi}{2}\right)\right) = -\frac{1}{2} \left(-1 - 0\right) = \frac{1}{2} \]

6. **Calculate the Second Integral**:
Evaluating from \( \frac{\pi}{2} \) to \( \frac{3\pi}{4} \):
\[ \left[-\frac{1}{2} \cos(2x)\right]_{\frac{\pi}{2}}^{\frac{3\pi}{4}} = -\frac{1}{2} \left(\cos\left(\frac{3\pi}{2}\right) - \cos(\pi)\right) = -\frac{1}{2} \left(0 - (-1)\right) = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

7. **Combine the Areas**:
The total area is:
\[ A = \frac{1}{2} + \frac{1}{2} = 1 \text{ square unit} \]

#### Part (ii): Area Between the Curve \( y = \cos(3x) \) and the Axes

1. **Determine the Period of the Function**:
The function \( y = \cos(3x) \) has a period given by:
\[ T = \frac{2\pi}{3} \]

2. **Identify the Limits**:
We need to find the area from \( x = 0 \) to \( x = \frac{\pi}{6} \).

3. **Sketch the Graph**:
The graph of \( y = \cos(3x) \) oscillates between -1 and 1. At \( x = 0 \), \( y = 1 \) and at \( x = \frac{\pi}{6} \), \( y = \cos\left(\frac{\pi}{2}\right) = 0 \).

4. **Set Up the Integral**:
The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{\frac{\pi}{6}} \cos(3x) \, dx \]

5. **Calculate the Integral**:
\[ \int \cos(3x) \, dx = \frac{1}{3} \sin(3x) \]
Evaluating from \( 0 \) to \( \frac{\pi}{6} \):
\[ \left[\frac{1}{3} \sin(3x)\right]_{0}^{\frac{\pi}{6}} = \frac{1}{3} \left(\sin\left(\frac{\pi}{2}\right) - \sin(0)\right) = \frac{1}{3} (1 - 0) = \frac{1}{3} \]

### Final Answers
- The area enclosed by \( y = \sin(2x) \) from \( x = \frac{\pi}{4} \) to \( x = \frac{3\pi}{4} \) is **1 square unit**.
- The area between the curve \( y = \cos(3x) \) and the axes from \( x = 0 \) to \( x = \frac{\pi}{6} \) is **\(\frac{1}{3}\) square units**.