Given x = sqrt(2) + sqrt(3) and y = sqrt(2) − sqrt(3).

$$x^2 = (\sqrt{2} + \sqrt{3})^2$$ $$x^2 = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2$$ $$x^2 = 2 + 2\sqrt{6} + 3$$ $$x^2 = 5 + 2\sqrt{6}$$

$$y^2 = (\sqrt{2} - \sqrt{3})^2$$ $$y^2 = (\sqrt{2})^2 - 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2$$ $$y^2 = 2 - 2\sqrt{6} + 3$$ $$y^2 = 5 - 2\sqrt{6}$$

$$x^2 + y^2 = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$$ $$x^2 + y^2 = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}$$ $$x^2 + y^2 = 10$$

The statement $x^2 + y^2 = 4$ is incorrect for the given values of $x$ and $y$. Instead, $\boxed{x^2 + y^2 = 10}$.

$$xy = (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})$$

This is a difference of squares, $(a+b)(a-b) = a^2 - b^2$.

$$xy = (\sqrt{2})^2 - (\sqrt{3})^2$$ $$xy = 2 - 3$$ $$xy = -1$$

The statement $xy = 1$ is incorrect for the given values of $x$ and $y$. Instead, $\boxed{xy = -1}$.