Degree of algebraic numbers

Minimal polynomial. $X^2-2$ is monic, irreducible over $\mathbb{Q}$ (rational root test), and annihilates $\sqrt2$. So it is the minimal polynomial.

Degree. $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=\deg(X^2-2)=2$.

Basis. Every element of $\mathbb{Q}(\sqrt2)$ is of the form $a+b\sqrt2$ with $a,b\in\mathbb{Q}$ (from arithmetic of $a+b\sqrt2$ and the inverse formula). So $\{1,\sqrt2\}$ spans; it is also independent (since $\sqrt2\notin\mathbb{Q}$).

Conclusion. Degree $2$, basis $\{1,\sqrt2\}$.