Abstract Algebra automorphisms inner automorphisms - Are my answers correct? Hope you are well.
Im posting the following as a favor on behalf of a friend who is not a native speaker, she just wants to confirm her answer is indeed correct.
Image of Questions:

Answer of 1st Question.

Answer of 2nd Question:

If possible please post any answers in the most simple form for easy understanding.
Any input would be greatly appreciated, Thanks in advance!
 A: Both questions seem fairly vague. For (ii), I'll assume that $(\mathbb{Q},+)$ refers to the underlying additive group of $\mathbb{Q}$. In that case, $\operatorname{Inn}(\mathbb{Q})$ refers specifically to those automorphisms generated by conjugations, but since $\mathbb{Q}$ is abelian, conjugations are trivial, and $\operatorname{Inn}(\mathbb{Q})$ is therefore trivial as well. Contrast this with $\operatorname{Aut}(\mathbb{Q})$: For every nonzero rational $q$, multiplication by $q$ is a nontrivial group automorphisms. This results in a homomorphism $\mathbb{Q}^{\times} \to \operatorname{Aut}(\mathbb{Q})$. In fact, this map is an isomorphism. To prove this, argue that, for any automorphism $f$, one has $f(n) = n\cdot f(1)$ for all integers $n$, and $f(1/n) = f(1) / n$ for all (nonzero) $n$ as well.
As for question (iii), I agree with lulu that the question is not clear. When they say 'group with multiplication', do they just mean the ring, and if so, is $\operatorname{Aut}(\mathbb{Q}[\sqrt{2}])$ the group of ring automorphisms? In that case, the provided answer seems correct: the underlying $\mathbb{Q}$ can only be mapped to itself trivially, and it brings us to the question what happens to $\sqrt{2}$ under an automorphism. Well, $\sqrt{2}$ is a root of $x^2 - 2$, and so must $f(\sqrt{2})$ be for any automorphism $f$, and there are two such roots, namely $\pm \sqrt{2}$. Thus, $\operatorname{Aut}(\mathbb{Q}[\sqrt{2}])$ consists of two elements, namely the identity and $a + b\sqrt{2} \mapsto a - b\sqrt{2}$.
