Vector spaces and subspaces. 
a) Find an abelian group V and a field F for which V is a vctor space over F in at least two different ways, that is, there are two different definitions of scalar multiplication making V a vector spae over F. 

We can choose both V and F to be $\Bbb{Q}$ and define the two different scalar multiplications by $aq := aq$ and $aq := a^{-1}q$ for all $a,q \in \Bbb{Q}$. Is this correct?

b) Find a vector space V over F and a subset S of V that is (1) a subspace of V and (2) a vector space using operations that differ from those of V. 

I find this a bit surprising, because the textbook (Advanced Linear Algebra by Roman) says "A subspace of a vector space V is a subset S of V that is a vector space in its own right under the operations obtained by restricting the operations of V to S...". However, if S has a different operation that V does, then it won't fit this definition, right? 
Thank you in advance
 A: Your answer to (a) doesn't work because $a^{-1}$ doesn't exist when $a=0$.  Even apart from this difficulty, your proposed multiplication won't satisfy the distributive law (with respect to scalar addition).  There won't be any example over the field of rational numbers; I suggest trying the field of complex numbers.
For (b), I think the intention is that $S$ should be a subspace of $V$ using (as the definition requires) the restrictions to $S$ of the operations of $V$, and should also be a vector space with some other operations.  In other words, it's similar to question (a) in that the same set is a vector space for two different choices of operations.
A: Your suggestion to part a) is wrong. The axiom
$$
(a+b)q=aq+bq
$$
does not hold. Try any field with a non-trivial automorphism. For example complex numbers acting on themselves either by multiplication or by multiplication with the conjugate.
For part b) let $S=F=\mathbb{Q}(\sqrt2)$ be the subspace of the $F$-space $\mathbb{R}$. The field $F$ has a non-trivial automorphism, which can be used to give it another structure as a vector space over $F$.
A: For part (a), a vector space over $F$ is an $F$-module, so to find a vector space over $F$ in at least two ways, you need to find an abelian group $M$, a field $F$ and at least two ring homomorphisms
$$F \to \operatorname{End}(M)$$
where $\operatorname{End}(M)$ is the endomorphism ring of $M$.   So if you already know a vector space $V$ over $F$, any non-identity automorphism of $F$ gives you a new vector space with the same field acting differently.
