Solution: Let $x = \sqrt{2} + \sqrt{3} $. Then we have \begin{align*} x^2 = 2 + 3 + 2\sqrt{6} & \implies x^2 - 5 = 2\sqrt{6} \\ & \implies (x^2 - 5)^2 = 24 \\ & \implies x^4 -10x^2 + 25 - 24 = 0 \\ & \implies x^4 - 10x^2 + 1 = 0. \end{align*} Since the polynomial $p(x) = x^4 - 10x^2 + 1 \in \mathbb{Q} [x]$, the number $\sqrt{2} + \sqrt{3} $ is algebraic over $\mathbb{Q} $.

To determine the degree $\sqrt{2} + \sqrt{3} $ over $\mathbb{Q} $, we observe that the polynomial $x^4 - 10x^2 + 1$ is irreducible over $\mathbb{Q} $. To see this note that the only possible rational roots are $\pm 1$. Since $p(1)$ and $p(-1)$ are nonzero, the polynomial $p(x)$ does not have a rational root. Also, $p(x)$ cannot be factored as a product of two quadratic polynomials. For that, let \begin{align*} & p(x) = \left( x^2 + ax + b\right) \left( x^2 - ax + b\right) \\ \implies & x^4 +x^2 \left( 2b - a^2 \right) + b^2. \end{align*} Comparing the coefficients, we get \begin{align*} b^2 = 1 \text{ and } 2b - a^2 = -10, \end{align*} the above does not have a solution in $\mathbb{Q} $. Thus, $p(x)$ is irreducible and hence $\left[ \mathbb{Q} \left( \sqrt{2} + \sqrt{3} : \mathbb{Q} \right) \right] = 4$.

Similarly to determine the degree of $\sqrt{2} \cdot \sqrt{3} $, we observe that if \[ x = \sqrt{2} \cdot \sqrt{3} \implies x^2 = 6. \] Thus, $x^2 - 6$ is the minimal polynomial (as the polynomial is irreducible over $\mathbb{Q} $) that satisfies $\sqrt{2} \cdot \sqrt{3} = \sqrt{6} $. Hence, $\left[ \mathbb{Q} (\sqrt{6} ) : \mathbb{Q} \right] = 2 $.