I'm having somewhat of a difficult time understand what's being asked—and thus having a hard time answering the question:
Find all conjugates in $\mathbb{C}$ of the given number over the given field:
1) $\sqrt{2}$ over $\mathbb{Q}$
2) $\sqrt{2}$ over $\mathbb{R}$
...
4) $\sqrt{2} - \sqrt{3}$ over $\mathbb{Q}$
...
7) $\sqrt{1 + \sqrt{2}}$ over $\mathbb{Q}$
Now: This means, in the first example, I must first find the monic irreducible polynomial of least degree in $\mathbb{Q}$ such that $\sqrt{2}$ is its zero, right? Is it right to go about doing this by saying "We know $\text{deg}( \text{irr}(\sqrt{2}, \mathbb{Q})) = 2$, so it has the form $x^2 + a_1x + a_2 = 0$ for $a_i \in \mathbb{Q}$ when $x = \sqrt{2}$. $a_2$ must be $-2$ and $a_1$ must be zero.
We have the polynomial $x^2 - 2$, for which $\sqrt{2}, -\sqrt{2}$ are both zeros, to the answer to this is $\sqrt{2}, -\sqrt{2}$."
In the case of$\sqrt{2}$ over $\mathbb{R}$, isn't $\sqrt{2} \in \mathbb{R}$—and isn't every polynomial reducible in $\mathbb{R}$?
For the third example, can I follow the same idea that I did in the first? Overall, I'm pretty confused about how to do these. I think there are a lot of holes in my understanding; I get what the question is asking for, but I'm not sure how to approach finding an answer.
So, you say $\sqrt2$ has no conjugate over $\Bbb R$ but has $-\sqrt2$ over $\Bbb Q$, which is correct.
For the other two, you just have to write up an equation that uses rationals for the given numbers, e.g. for 4) $$\begin{align}x&=\sqrt2-\sqrt3 \\ x+\sqrt3&=\sqrt2\\x^2+2\sqrt3x+3&=2 \\ x^2+1&= - 2\sqrt3 x\\ \dots&\dots \end{align}$$ From this you can also track back the other roots of the final equation.