$$\cos \left(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{33}{65}\right)$$ is equal to:

$$\alpha = \sin^{-1}\frac{3}{5}$$ $$\beta = \sin^{-1}\frac{5}{13}$$ $$\gamma = \sin^{-1}\frac{33}{65}$$

From these, we can find the sine and cosine values for each angle since they lie in the first quadrant:

$$\sin\alpha = \frac{3}{5}, \quad \cos\alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$$ $$\sin\beta = \frac{5}{13}, \quad \cos\beta = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$$ $$\sin\gamma = \frac{33}{65}, \quad \cos\gamma = \sqrt{1 - \left(\frac{33}{65}\right)^2} = \frac{56}{65}$$

Next, we can calculate $$\sin(\alpha + \beta)$$ and $$\cos(\alpha + \beta)$$ using the trigonometric sum formulas:

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) = \frac{36 + 20}{65} = \frac{56}{65}$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) = \frac{48 - 15}{65} = \frac{33}{65}$$

Now, we rewrite the original expression as $$\cos((\alpha + \beta) + \gamma)$$ and expand it:

$$\cos(\alpha + \beta + \gamma) = \cos(\alpha + \beta)\cos\gamma - \sin(\alpha + \beta)\sin\gamma$$

Substitute the known values into the equation:

$$\cos(\alpha + \beta + \gamma) = \left(\frac{33}{65}\right)\left(\frac{56}{65}\right) - \left(\frac{56}{65}\right)\left(\frac{33}{65}\right) = 0$$

Thus, the value of the expression is:$$0$$