Question: I haven't taken any abstract algebra class yet, and for those linear algebra classes I did take, the discussions are mainly focusing on real and complex vector spaces. From textbooks and other resources, I noticed that, in principle, it is allowed to have vector spaces over other fields. So I tried to make up some random examples, but soon I found that the situation could become cumbersome quite easily:

We could definitely have $\mathbb{C}^n$ as a vector space over $\mathbb{C}$.

So I think it should also be fine to have $\mathbb{C}^n$ as a vector space over $\mathbb{R}$, and I checked all the properties for it to be a vector space, it seems fine indeed.

However, if we have something like $\mathbb{R}^n$ over $\mathbb{C}$, things are getting weird (as the scalar multiplication with usual rule can now take the vector out from $\mathbb{R}^n$), so I tried to modify the definition of the scalar multiplication, but it seems quite difficult (and useless) to turn such structure into a vector space.

I've also tried some other examples, including $\mathbb{R}^n$ over {0,1} where {0,1} is the field with arithmetic modulo 2. Things are getting worse here as the arithmetics for the field {0,1} are not even the same as for $\mathbb{R}$.

So I wonder, apart from explicitly checking the arithmetic properties for a structure to be vector space every time, if there are some restrictions or requirements that are needed to be satisfied between a vector space and the field containing its scalars?

Answer: 
Clearly, $f(x)$ for some $x$ implies $f(x)=0$ for all $x$. Excluding this case the question can be translated to Cauchy's equation : $g(x+y)=g(x)+g(y)$ by taking logarithm. Here are some facts about $g$: If $g$ is Borel measurable (in particular if it is monotone) then $g(x)=cx$ for some constant $c$. But there exist non-measurable solution so of this equation. [Proof of the existence of such functions requires Axiom of Choice].

Hence $f(x)=a^{x}$ need not be true in general (take $f(x)=e^{g(x)})$.
**No.** There exist non-vanishing such functions which are not exponential.

First observe that, if $g(x+y)=g(x)+g(y)$, for all $x,y$, and $f(x)=\exp\big(g(x)\big)$, then $f(x+y)=f(x)f(y)$, for all $x,y$.

To obtain an additive $g$, i.e., $g(x+y)=g(x)+g(y)$, for all $x,y$, which is not of the form $g(x)=cx$, we need to use Zorn's Lemma, and in particular, the fact that $\mathbb R$ possesses a Hamel basis, as a linear space over $\mathbb Q$. In other words, there exists a $B\subset \mathbb R$, such that:

Every $x\in\mathbb R$ can be written, in a unique way as a linear combination of elements of $B$ with rational coefficients. That is, for every $x\in\mathbb R$, there exist unique $b\_1,\ldots,b\_n\in B,$ and $q\_1,\ldots,q\_n\in\mathbb Q$, such that
$$
x=q\_1b\_1+\cdots+q\_nb\_n.
$$

Now, for every set of $c\_i$'s in $\mathbb R$, the function
$$
g(x)=q\_1q(b\_1)+\cdots+q\_ng(b\_n)=q\_1c\_1+\cdots+q\_nc\_n,
$$
is additive, i.e., $g(x+y)=g(x)+g(y)$, and the coresponding $f$, i.e., $f(x)=e^{g(x)}$ satisfies
$$
g(x+y)=g(x)g(y).
$$
If the $c\_i$'s are not proportional to the $b\_i$'s, then $f$ is not continuous and hence not exponential. For example, say $g(b\_i)=c\_i$, $i=1,2$, and $b\_1c\_2-b\_2c\_1\ne 0$. Say $c\_1\ne 0$, and $\{q\_n\}\in \mathbb Q$, so that $q\_n\to b\_1/b\_2$. Then
$$
f(q\_nb\_2)=q\_nc\_2\to b\_1c\_2/b\_2
$$
while
$$
q\_n b\_2\to b\_2\cdot b\_1/b\_2=b\_1\quad\text{and}\quad f(b\_1)=c\_1\ne b\_1c\_2/b\_2.
$$