Let $$x = \frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\cos\left(2\sin^{-1}x\right) = \frac{1}{9}$$ and let $$\alpha, \beta (\alpha > \beta)$$ be the roots of the equation $$mx^2 - nx - m + n = 0$$. Then the point $$(\alpha, \beta)$$ lies on the line

$$\cos(2\sin^{-1}x) = 1 - 2x^2$$.

$$1 - 2x^2 = \frac{1}{9}$$, so $$2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$$ and hence $$x^2 = \frac{4}{9}$$ which implies $$x = \frac{2}{3}$$. Because $$x$$ is positive, we conclude $$m = 2, n = 3$$.

 $$mx^2 - nx - m + n = 0$$.

Substituting $$m=2$$ and $$n=3$$ $$2x^2 - 3x - 2 + 3 = 0$$ or $$2x^2 - 3x + 1 = 0$$. 

Factoring gives $$(2x - 1)(x - 1) = 0$$, so the roots are $$x = \frac{1}{2}$$ and $$x = 1$$. 

Since we denote the larger root by $$\alpha$$ and the smaller by $$\beta$$, we have $$\alpha = 1$$ and $$\beta = \frac{1}{2}$$.

Finally, we check which line passes through the point $$(1, \frac{1}{2})$$.

Substituting into Option (4): $$5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9$$,