Solve the equation:
(i) `(7x-5)/(8x+3)gt4`
(ii) `(x)/(x-5)gt(1)/(2)`.

Answer

Let's solve the given inequalities step by step.

### Part (i): Solve the inequality \(\frac{7x - 5}{8x + 3} > 4\)

1. **Rearranging the Inequality**:
We start by rewriting the inequality:
\[ \frac{7x - 5}{8x + 3} - 4 > 0 \]
This can be expressed with a common denominator:
\[ \frac{7x - 5 - 4(8x + 3)}{8x + 3} > 0 \]

2. **Simplifying the Numerator**:
Now, simplify the numerator:
\[ 7x - 5 - 32x - 12 = -25x - 17 \]
So, the inequality becomes:
\[ \frac{-25x - 17}{8x + 3} > 0 \]

3. **Finding Critical Points**:
We need to find the critical points where the numerator and denominator are zero:
- For the numerator:
\(-25x - 17 = 0 \Rightarrow x = -\frac{17}{25}\)
- For the denominator:
\(8x + 3 = 0 \Rightarrow x = -\frac{3}{8}\)

4. **Number Line and Test Intervals**:
We now plot these critical points on a number line:
\[ -\frac{17}{25} \approx -0.68 \quad \text{and} \quad -\frac{3}{8} \approx -0.375 \]
The intervals to test are:
\((- \infty, -\frac{17}{25})\), \((- \frac{17}{25}, -\frac{3}{8})\), and \((- \frac{3}{8}, \infty)\).

5. **Testing Intervals**:
- For \(x < -\frac{17}{25}\) (e.g., \(x = -1\)):
\[ \frac{-25(-1) - 17}{8(-1) + 3} = \frac{25 - 17}{-8 + 3} = \frac{8}{-5} < 0 \]
- For \(-\frac{17}{25} < x < -\frac{3}{8}\) (e.g., \(x = -0.5\)):
\[ \frac{-25(-0.5) - 17}{8(-0.5) + 3} = \frac{12.5 - 17}{-4 + 3} = \frac{-4.5}{-1} > 0 \]
- For \(x > -\frac{3}{8}\) (e.g., \(x = 0\)):
\[ \frac{-25(0) - 17}{8(0) + 3} = \frac{-17}{3} < 0 \]

6. **Conclusion for Part (i)**:
The solution is where the expression is positive:
\[ x \in \left(-\frac{17}{25}, -\frac{3}{8}\right) \]

### Part (ii): Solve the inequality \(\frac{x}{x - 5} > \frac{1}{2}\)

1. **Rearranging the Inequality**:
Rewrite the inequality:
\[ \frac{x}{x - 5} - \frac{1}{2} > 0 \]
This can be expressed with a common denominator:
\[ \frac{2x - (x - 5)}{2(x - 5)} > 0 \]
Simplifying gives:
\[ \frac{x + 5}{2(x - 5)} > 0 \]

2. **Finding Critical Points**:
We find the critical points where the numerator and denominator are zero:
- For the numerator:
\(x + 5 = 0 \Rightarrow x = -5\)
- For the denominator:
\(2(x - 5) = 0 \Rightarrow x = 5\)

3. **Number Line and Test Intervals**:
We plot these critical points on a number line:
\(-5\) and \(5\).
The intervals to test are:
\((- \infty, -5)\), \((-5, 5)\), and \((5, \infty)\).

4. **Testing Intervals**:
- For \(x < -5\) (e.g., \(x = -6\)):
\[ \frac{-6 + 5}{2(-6 - 5)} = \frac{-1}{-22} > 0 \]
- For \(-5 < x < 5\) (e.g., \(x = 0\)):
\[ \frac{0 + 5}{2(0 - 5)} = \frac{5}{-10} < 0 \]
- For \(x > 5\) (e.g., \(x = 6\)):
\[ \frac{6 + 5}{2(6 - 5)} = \frac{11}{2} > 0 \]

5. **Conclusion for Part (ii)**:
The solution is where the expression is positive:
\[ x \in (-\infty, -5) \cup (5, \infty) \]

### Final Answers:
1. For part (i): \(x \in \left(-\frac{17}{25}, -\frac{3}{8}\right)\)
2. For part (ii): \(x \in (-\infty, -5) \cup (5, \infty)\)