Let $$\alpha, \beta$$ be the roots of the equation $$x^2 - \sqrt{2}x + \sqrt{6} = 0$$ and $$\frac{1}{\alpha^2+1}, \frac{1}{\beta^2+1}$$ be the roots of the equation $$x^2 + ax + b = 0$$. Then the roots of the equation $$x^2 - (a+b-2)x + (a+b+2) = 0$$ are:

We are given that $$\alpha, \beta$$ are roots of $$x^2 - \sqrt{2}x + \sqrt{6} = 0$$. By Vieta's formulas, $$\alpha + \beta = \sqrt{2}$$ and $$\alpha\beta = \sqrt{6}$$.

Now $$\frac{1}{\alpha^2+1}$$ and $$\frac{1}{\beta^2+1}$$ are the roots of $$x^2 + ax + b = 0$$. We compute their sum and product.

First, $$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 2 - 2\sqrt{6}$$.

The product of the denominators is $$(\alpha^2+1)(\beta^2+1) = (\alpha\beta)^2 + (\alpha^2 + \beta^2) + 1 = 6 + (2 - 2\sqrt{6}) + 1 = 9 - 2\sqrt{6}$$.

So the sum of the new roots is:

And the product is:

We need $$a + b = \frac{2\sqrt{6} - 4 + 1}{9 - 2\sqrt{6}} = \frac{2\sqrt{6} - 3}{9 - 2\sqrt{6}}$$. Rationalizing:

The final equation is $$x^2 - (a+b-2)x + (a+b+2) = 0$$. With $$s = a + b = \frac{4\sqrt{6}-1}{19}$$:

The sum of roots $$= s - 2 = \frac{4\sqrt{6} - 39}{19}$$. Since $$4\sqrt{6} \approx 9.8$$, this is approximately $$\frac{-29.2}{19} < 0$$.

The product of roots $$= s + 2 = \frac{4\sqrt{6} + 37}{19} > 0$$.

When the sum of two roots is negative and their product is positive, both roots must be negative (since two numbers with a positive product have the same sign, and their negative sum forces both to be negative).

Hence, the correct answer is Option 2.