Find degree $[\mathbb{Q}(i,\sqrt{2}) : \mathbb{Q}]$
let $a = \sqrt{2}$
$a^2 = 2$
$\therefore a$ is a root of $q(x) = x^2 -2$, where $q(x)\in\mathbb{Q}(i\sqrt{2})[x]$
means degree of $a$ over $\mathbb{Q}(i\sqrt{2})$ is $\leq 2$
If degree is $1$ then $a\in\mathbb{Q}(i\sqrt{2})$, which clearly not possible as $\mathbb{Q}(i\sqrt{2}) = [{a+bi\sqrt{2}|a,b,\in\mathbb{Q}}]$
$\therefore$ Degree of $[\mathbb{Q}(i,\sqrt{2}) : \mathbb{Q}(i\sqrt{2})]$ is $2$
Does this makes sense?
For the second one, remember that we're thinking of $\mathbb Q(i, \sqrt 2)$ as an extension of $Q(i\sqrt2)$, not of $\mathbb Q$. Writing $K=\mathbb Q(i\sqrt 2)$, what can you adjoin to $K$ to get $\mathbb Q(i, \sqrt 2)$? Then you just need to look for the minimal polynomial of that element over $K$.
In fact, you can show that it suffices to adjoin either $i$ or $\sqrt 2$ to $K$. You can find the minimum polynomial of either of those over $K$ the same way you would find their minimum polynomial over $\mathbb Q$.
To show that $K(i)=\mathbb Q(i, \sqrt2)$ (for example), use the definition that, when $F$ is any field and $a$ is an element of an extension of $F$, $F(a)$ is the smallest (by inclusion) extension of $F$ that contains $a$. So $K(i)$ is the smallest extension of $K$ that contains $i$. Well, $\mathbb Q(i, \sqrt2)$ is an extension of $K$ that contains $i$, so therefore $K(i)\subseteq\mathbb Q(i, \sqrt2)$. But $K(i)$ also contains both $i$ and $\sqrt2$ (since $K$ contains $i\sqrt2$, so just divide that by $i$). So it's an extension of $\mathbb Q$ containing both $i$ and $\sqrt2$, and therefore contains the smallest such extension $\mathbb Q(i, \sqrt2)$. So $K(i)\supseteq\mathbb Q(i, \sqrt2)$.