With practice, one learns to recognize the sort of things that may go wrong with potential "vector spaces", and quickly zoom in on those. But, the thing is, it takes practice to figure this out.
Often, if one thing goes wrong, lots of things will go wrong; sometimes, it is one and only one thing that goes wrong (and it may be hard to spot). At this stage, it might actually be a good idea for you to check each axiom and see whether it is met or not met, because it will afford you a lot of practice. Even though it's enough to find one axiom that fails for something to not be a vector space, finding all the ways in which things go wrong is likely good practice at this stage.
For example, you don't say which problem "says the answer is Axiom 4", and in fact I see no problem, among the ones listed, in which $4x+1$ is even a vector! It's not a $4\times 6$ matrix, it's not a $1\times 1$ matrix, it's not a degree 3 polynomial, it's not a degree 5 polynomial, it's not a first degree polynomial whose graph passes through the origin, and it's not a quadratic function whose graph passes through the origin...
Since user6312 already got you started with Question 15, let's continue: you know it fails Axiom 1. It is not hard to verify that it satisfies axioms 2 and 3. Axiom 4 fails because the zero vector (the polynomial 0) is not in your set....
Axiom 5 is a bit tricky: strictly speaking, Axiom 5 does not even make sense if Axiom 4 fails, because there is no $\mathbf{0}$ in the first place. I would certainly score such a statement as correct. On the other hand, if you have a polynomial of degree exactly 3, $ax^3+bx^2+cx+d$, with $a\neq 0$, then you can find a polynomial of degree exactly 3 that added to it will give you the zero polynomial (which is not in the set). So you might also say Axiom 5 is "sort of" satisfied.
Axiom 6 fails: for example, $x^3$ is in your set, $c=0$ is a scalar, but $0(x^3)$ is not in your set.
It's not hard to verify that Axioms 7, 8, 9, and 10 do hold.
So for 15, the axioms that fail are Axioms 1, 4, 6, and possibly 5 (depending how you interpret it).
You'll find similar problems with 16. There's a bit more to do with 17, because you also have the condition "and passes through the origin"; be sure to take that into account. Similar with 18. As for 13 and 14, I'll spill the beans and tell you that they are vector spaces: you should verify that all the axioms hold, one by one. Be sure to not verify them "by example": it's not enough to show that for particular $4\times 6$ matrices $\mathbf{u}$ and $\mathbf{v}$ you have $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$: you must verify it works for all possible choices of $\mathbf{u}$ and $\mathbf{v}$. If you find yourself saying "since, for example..." chances are you're doing it wrong.
You are mistaking "pointwise scalar multiplication" for "scalar product". For $\alpha \in \Bbb R$ and $\mathbf x \in \mathbf G$ one may define the poinwise scalar multiplication by $(\alpha \mathbf x) (\omega) = \alpha \big( \mathbf x (\omega) \big)$, for $\omega \in \Omega$. There is no mention of scalar product here.
Of course, if you really wanted, you could also define a scalar product by $\langle \mathbf x, \mathbf y \rangle = \int _\Omega \mathbf x (\omega) \mathbf y (\omega) \ \Bbb d \mu$ where $\mu$ is the measure on $\Omega$, but care must be taken in order to assure the convergence of the above integral for all $\mathbf x$ and $\mathbf y$, and in order to make this scalar product non-degenerate (you'll have to work with classes of functions etc.). This could be done, for instance, by requiring $\mu$ to be finite. In any case, this is not what your text tries to convey.
Best Answer
For any subset $A$ of $\Omega$ define $0\cdot A=\emptyset,\,1\cdot A=A$. For any subsets $A,\,B$ define $A+B$ as the disjoint union of $A,\,B$ (this requires the convention $1+1=0$). The singletons form a basis of the space, with $A$ the sum of the singletons that are subsets of it.