Solve :
`(i)" "tan^(-1). x/2 +tan^(-1). x/3 = pi/4 , sqrt(6) gt x gt 0 `
`(ii)" "sin(sin^(-1) . 1/5+cos^(-1)x)=1`

Answer

Let's solve the given problems step by step.

### Part (i): Solve \( \tan^{-1} \left( \frac{x}{2} \right) + \tan^{-1} \left( \frac{x}{3} \right) = \frac{\pi}{4} \)

1. **Use the identity for the sum of inverse tangents**:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
Here, let \( a = \frac{x}{2} \) and \( b = \frac{x}{3} \).
\[ \tan^{-1} \left( \frac{x}{2} \right) + \tan^{-1} \left( \frac{x}{3} \right) = \tan^{-1} \left( \frac{\frac{x}{2} + \frac{x}{3}}{1 - \frac{x}{2} \cdot \frac{x}{3}} \right) \]

2. **Calculate \( \frac{x}{2} + \frac{x}{3} \)**:
\[ \frac{x}{2} + \frac{x}{3} = \frac{3x + 2x}{6} = \frac{5x}{6} \]

3. **Calculate \( 1 - \frac{x}{2} \cdot \frac{x}{3} \)**:
\[ 1 - \frac{x}{2} \cdot \frac{x}{3} = 1 - \frac{x^2}{6} = \frac{6 - x^2}{6} \]

4. **Substitute back into the identity**:
\[ \tan^{-1} \left( \frac{\frac{5x}{6}}{\frac{6 - x^2}{6}} \right) = \tan^{-1} \left( \frac{5x}{6 - x^2} \right) \]

5. **Set the equation equal to \( \tan \frac{\pi}{4} \)**:
\[ \tan^{-1} \left( \frac{5x}{6 - x^2} \right) = \frac{\pi}{4} \implies \frac{5x}{6 - x^2} = 1 \]

6. **Cross-multiply to solve for \( x \)**:
\[ 5x = 6 - x^2 \]
Rearranging gives:
\[ x^2 + 5x - 6 = 0 \]

7. **Factor the quadratic equation**:
\[ (x + 6)(x - 1) = 0 \]

8. **Find the values of \( x \)**:
\[ x + 6 = 0 \implies x = -6 \quad \text{(not valid since } x > 0\text{)} \]
\[ x - 1 = 0 \implies x = 1 \]

9. **Check the range**:
Since \( 0 < x < \sqrt{6} \) and \( \sqrt{6} \approx 2.45 \), \( x = 1 \) is valid.

### Solution for Part (i):
\[ \boxed{1} \]

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### Part (ii): Solve \( \sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \)

1. **Recognize that \( \sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2} \)**:
\[ \sin(\sin^{-1} a + \cos^{-1} b) = 1 \implies \sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2} \]

2. **Rearranging gives**:
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} \frac{1}{5} \]

3. **Use the identity \( \cos^{-1} a = \sin^{-1} \sqrt{1 - a^2} \)**:
\[ x = \sin\left(\frac{\pi}{2} - \sin^{-1} \frac{1}{5}\right) = \cos(\sin^{-1} \frac{1}{5}) \]

4. **Calculate \( \cos(\sin^{-1} \frac{1}{5}) \)**:
Using the Pythagorean identity:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \]
where \( \sin(\theta) = \frac{1}{5} \):
\[ \cos(\sin^{-1} \frac{1}{5}) = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \]

### Solution for Part (ii):
\[ \boxed{\frac{2\sqrt{6}}{5}} \]

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