If α, β are two different values of x lying between 0 and 2π,

Given:

$6 \cos x+8 \sin x=9$

$\Rightarrow 6 \cos x=9-8 \sin x$

$\Rightarrow 36 \cos ^{2} x=(9-8 \sin x)^{2}$

$\Rightarrow 36\left(1-\sin ^{2} x\right)=81+64 \sin ^{2} x-144 \sin x$

$\Rightarrow 100 \sin ^{2} x-144 \sin x+45=0$

Now, $\alpha$ and $\beta$ are the roots of the given equation; therefore, $\cos \alpha$ and $\cos \beta$ are the roots of the above equation.

$\Rightarrow \sin \alpha \sin \beta=\frac{45}{100} \quad$ (Product of roots of a quadratic equation $a x^{2}+b x+c=0$ is $\frac{c}{a} .$ )

Again, $6 \cos x+8 \sin x=9$

$\Rightarrow 8 \sin x=9-6 \cos x$

$\Rightarrow 64 \sin ^{2} x=(9-6 \cos x)^{2}$

$\Rightarrow 64\left(1-\cos ^{2} x\right)=81+36 \cos ^{2} x-108 \cos x$

$\Rightarrow 100 \cos ^{2} x-108 \cos x+17=0$

Now, $\alpha$ and $\beta$ are the roots of the given equation; therefore, $\sin \alpha$ and $\sin \beta$ are the roots of the above equation.

Therefore, $\cos \alpha \cos \beta=\frac{17}{100}$

Hence, $\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$

$=\frac{17}{100}-\frac{45}{100}$

$=-\frac{28}{100}$

$=-\frac{7}{25}$

$\sin (\alpha+\beta)=\sqrt{1-\cos ^{2}(\alpha+\beta)}$

$=\sqrt{1-\left(\frac{-7}{25}\right)^{2}}$

$=\sqrt{\frac{576}{625}}$

$=\frac{24}{25}$