what is the meaning of $\mathbb{Q}(\sqrt{2},i)$

$\mathbb Q(\sqrt 2,i)$ is the smallest field that contains $\mathbb Q$ and $\sqrt 2 $ and $i$. The general element looks like $a+b\sqrt 2+ci+di\sqrt 2$ with $a,b,c,d\in\mathbb Q$.

Also note that in general $\mathbb Q(\xi)$ (where $\xi$ is some real or complex irrational number) need not be the set $\{\,a+b\xi\mid a,b\in\mathbb Q\,\}$. This is only true if $\xi$ is a quadratic irrational (such as $\sqrt 2$, $i$, or $\frac{1+\sqrt 5}2$), i.e. the root of a quadratic polynomial with rational coefficients. On the other hand $\mathbb Q(\sqrt[3]2)=\{\,a+b\sqrt[3]2+c\sqrt[3]4\mid a,b,c\in\mathbb Q\,\}$ and $\mathbb Q(\pi)=\{\,\frac{a_0+a_1\pi+\ldots +a_n\pi^n}{1+b_1\pi+\ldots +b_m\pi^m}\mid n,m\in\mathbb N, a_i,b_i\in\mathbb Q\,\}$.

For $F$ a field contained in another field $E$. For $x\in E$ it is common to let $F(x)$ denote the smallest field that contains both $F$ and $x$. For $x,y\in E$, $F(x,y)$ denotes the smallest field containing $F$, $x$, and $y$.

So for your concrete example, $\mathbb{Q}(\sqrt{2}, i)$ is the smallest extension of $\mathbb{Q}$ that contains $\sqrt{2}$ and $i$. It is not too hard to show that (as given in Hagen's answer) $$ \mathbb{Q}(\sqrt{2}, i) = \{a + b\sqrt{2} + ci + di\sqrt{2} \mid a,b,c,d\in \mathbb{Q}\}. $$ For example, we must have $\sqrt{2}$ and $i$ in the field, so we must have all elements of the form $a + b\sqrt{2} + ci$. We must also have the product of $\sqrt{2}$ and $i$, so the field must contain any element of the form $a + b\sqrt{2} + ci + di\sqrt{2}$. Now you just need to show that $$ \{a + b\sqrt{2} + ci + di\sqrt{2} \mid a,b,c,d\in \mathbb{Q}\} $$ is indeed a field. Is it closed under multiplication? Can you find an inverse to an element of this form?