1. Solve $\cos(2\sin^{-1}x)=\dfrac{1}{9}$

Let $\theta=\sin^{-1}x \implies \sin\theta=x.$ Using the identity $\cos 2\theta=1-2\sin^2\theta,$ we get $\cos\left(2\sin^{-1}x\right)=1-2x^2.$ So the equation becomes $1-2x^2=\frac{1}{9}.$

Thus, $2x^2=1-\frac{1}{9}=\frac{8}{9}$ $x^2=\frac{4}{9}$ $x=\pm \frac{2}{3}.$

Since $x=\dfrac{m}{n}$ with $m,n$ co-prime natural numbers, we must have $x=\frac{2}{3}.$ Hence, $m=2,\quad n=3.$

2. Form the quadratic whose roots are $\alpha,\beta$

Given equation: $mx^2-nx-m+n=0.$ Substituting $m=2, n=3$: $2x^2-3x-2+3=0$ $2x^2-3x+1=0.$

Factorizing, $2x^2-3x+1=(2x-1)(x-1)=0.$ So the roots are $x=1,\quad x=\frac{1}{2}.$

Given $\alpha>\beta$, we have $\alpha=1,\quad \beta=\frac{1}{2}.$ Thus the point is $(\alpha,\beta)=\left(1,\frac{1}{2}\right).$

3. Check which line passes through $\left(1,\frac{1}{2}\right)$

• A: $3x-2y=-2$ $3(1)-2\left(\frac{1}{2}\right)=3-1=2\ne -2$ Not correct.

• B: $3x+2y=2$ $3(1)+2\left(\frac{1}{2}\right)=3+1=4\ne 2$ Not correct.

• C: $5x+8y=9$ $5(1)+8\left(\frac{1}{2}\right)=5+4=9$ Correct.

• D: $5x-8y=-9$ $5(1)-8\left(\frac{1}{2}\right)=5-4=1\ne -9$ Not correct.

Therefore, the point $(\alpha,\beta)$ lies on $\boxed{5x+8y=9}.$

So the correct option is C.