$\mathbb{Q}$ is the field of rational numbers.
$\mathbb{Q}[x]$ is the ring of polynomials in one indeterminate $x$ with rational coefficients.
$\mathbb{Q}(x)$ is the field of fractions of $\mathbb{Q}[x]$; it is called the field of rational functions with coefficients in $\mathbb{Q}$. Its elements are of the form $\frac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomials with rational coefficients, and where
$$\frac{p(x)}{q(x)} = \frac{r(x)}{s(x)}\iff p(x)s(x)=q(x)r(x);$$
etc.
$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}$, $\sqrt{2}$, and $\sqrt{3}$. In principle, it will be equal to
$$\left.\left\{\frac{p(\sqrt{2},\sqrt{3})}{q(\sqrt{2},\sqrt{3})}\right| p(x,y),q(x,y)\in\mathbb{Q}[x,y], q(\sqrt{2},\sqrt{3})\neq 0\right\},$$
though in fact one can show that every element can be written uniquely as
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6},\qquad a,b,c,d\in\mathbb{Q}.$$
In general, if $F\subseteq K$ are fields, and $\alpha\in K$, then $F[\alpha]$ denotes the smallest subring of $K$ that contains $F$ and $\alpha$, and $F(\alpha)$ denotes the smallest subfield of $K$ that contains $F$ and $\alpha$. It is not hard to prove that, as sets,
$$\begin{align*}
F[\alpha] &= \{ p(\alpha)\mid p(x)\in F[x]\},\\
F(\alpha) &= \{p(\alpha)/q(\alpha)\mid p(x),q(x)\in F[x], q(\alpha)\neq 0\}.
\end{align*}$$
Though if $\alpha$ is algebraic over $F$, there are simpler expressions.
Similarly, if $S\subseteq K$ is a subset of $K$ (finite or infinite), then
$$\begin{align*}
F[S] &= \{ p(s_1,\ldots,s_n)\mid n\in\mathbb{N}, p(x_1,\ldots,x_n)\in F[x_1,\ldots,x_n], s_1,\ldots,s_n\in S\}\\
F(S) &= \left.\left\{ \frac{p(s_1,\ldots,s_n)}{q(s_1,\ldots,s_n)}\right| n\in\mathbb{N}, p,q\in F[x_1,\ldots,x_n], q(s_1,\ldots,s_n)\neq 0\right\}.
\end{align*}$$