Let sin-1(3/5) = x and sin-1(8/17) = y

Therefore  sinx = 3/5 and siny = 8/17

Now, cosx = √(1 - sin2x) = √(1 - (3/5)2) = √(1 - 9/25) = 4/5 and cosy = √(1 - sin2y) = √(1 - (8/17)2) = √(1 - 64/289) = 15/17

We have cos(x - y) = cosx cosy + sinx siny  = 4/5 x 15/17 + 3/5 x 8/17 = 60/85 + 24/85 = 84/85 ⇒ x - y = cos-1(84/85) ⇒ sin-1(3/5) - sin-1(8/17) = cos-1(84/85)

Let's denote \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) \) as \( \alpha \) and \( \tan^{-1} \left( \frac{x + 1}{x + 2} \right) \) as \( eta \).

Given that \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) + \tan^{-1} \left( \frac{x + 1}{x + 2} \right) = \frac{\theta}{4} \), we can use the tangent addition formula:

\[ \tan(\alpha + eta) = \frac{\tan \alpha + \tan eta}{1 - \tan \alpha \cdot \tan eta} \]

Substitute \( \tan \alpha = \frac{x - 1}{x - 2} \) and \( \tan eta = \frac{x + 1}{x + 2} \):

\[ \tan(\alpha + eta) = \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} \]

\[ \tan(\alpha + eta) = \frac{\frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{1 - \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}} \]

\[ \tan(\alpha + eta) = \frac{x^2 + x - 2 + x^2 - x - 2}{(x - 2)(x + 2) - (x^2 - 1)} \]

\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{x^2 + 4 - x^2 + 1} \]

\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{5} \]

Given that \( \tan(\alpha + eta) = \frac{\theta}{4} \), we have:

\[ \frac{2x^2 - 4}{5} = \frac{\theta}{4} \]

\[ 8x^2 - 16 = 5\theta \]

\[ 8x^2 = 5\theta + 16 \]

\[ x^2 = \frac{5\theta + 16}{8} \]

\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]

So, the value of \( x \) depends on the value of \( \theta \).

In LaTeX code:
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]