Elements of a real vectorspace certainly have direction, but they don't really have a magnitude. Well actually, they... kind-of have a magnitude. But for a proper magnitude, you need further structure, such as a norm or inner product. Let me explain.
Vector Spaces.
Suppose $V$ is a real vectorspace.
Definition 0. Given a vectors $x,y \in V$, we say that $x$ and $y$
have the same direction iff:
- there exists $r \in \mathbb{R}_{\geq 0}$ such that $x = ry,$ and
- there exists $r \in \mathbb{R}_{\geq 0}$ such that $y = rx$.
(The $r$'s don't have to be the same.)
This induces an equivalence relation on $V$, so we get a partitioning of $V$ into cells. Each cell is an open ray, so long as we regard $\{0\}$ as an open ray. You may wish to exclude $\{0\}$ from its privileged position as a ray, in which case you should only deal with non-zero vectors; that is, you need to be dealing with $V \setminus \{0\}$ rather than $V$.
Irrespective of which conventions are used, we can make sense of direction using these ideas:
Definition 1. The direction of $x \in V$ is the unique open ray $R \subseteq V$ such that $x \in R$.
Notice that the equivalence relation of having the same direction is preserved under scalar multiplication; what I mean is that if $v$ and $w$ have the same direction, then $av$ and $aw$ have the same direction, for any $a \in \mathbb{R}$. Geometrically, this means that if we scale a ray, we'll end up with a subset of another ray.
As for magnitude; well, if you choose a ray $R \subseteq V$, then we can partially order $R$ as follows. Given $x,y \in R$, we define that $x \geq y$ iff $x = ry$ for some $r \in \mathbb{R}_{\geq 1}$. So some vectors along this ray are longer than others, hence magnitude.
Inner Product Spaces.
Actually, this isn't the whole story. The problem with vector spaces is that if $x$ and $y$ don't belong to the same ray (nor to the the "negatives" of each others rays), then there's no way of comparing the magnitudes of $x$ and $y$. We can't say which is longer! Now there are mathematical situations where this limitation is desirable, but physically, you probably don't want this. A related issue is that you can't really make sense of angles in a (mere) vector space; at least, not without some further structure.
For this reason, when physicists say "vector", what they usually mean is "element of a finite-dimensional inner-product space." This is a (finite-dimensional) vector space $V$ with further structure; in particular, it comes equipped with a function
$$\langle-,-\rangle : V \times V \rightarrow \mathbb{R}$$
that is required to satisfy certain axioms resembling the dot product. Especially important for us is that these axioms include a "non-negativity" condition:
$$\langle x,x\rangle \geq 0$$
Using this, we can define the magnitude of vectors as follows.
Definition 2. Suppose $V$ is a real inner product space. Then the norm (or "magnitude") of $x \in V$, denoted $\|x\|$, is defined a follows:
$$\|x\| = \langle x,x\rangle^{1/2}$$
This allows us to compare the magnitudes of vectors that don't live in the same ray; we simply define that $x \geq y$ means $\|x\| \geq \|y\|.$ When confined to a single ray, this agrees with our earlier definition! Be careful though, because the relation $\geq$ we just defined is only a preorder.
In fact, the inner product gives us more than just magnitudes; it also gives angles!
Definition 3. Suppose $V$ is a real inner product space. Then the angle between of $x,y \in V$, denoted $\mathrm{ang}(x,y)$, is defined a follows:
$$\mathrm{ang}(x,y) = \cos^{-1}\left(\frac{\langle x,y\rangle}{\|x\|\|y\|}\right)$$
It can be shown that vectors $x$ and $y$ have the same direction (in the sense described at the beginning of my post) iff the angle between them is $0$. In fact, you can modify the above definition so that it defines the angle between any two non-zero open rays. In this case, it turns out that two rays are equal iff the angle between them is $0$.
Rough hints:
The zero vector is a vector $(x_0, y_0)$ such that for all $(x,y)\in Q^2$, we have $(x_0,y_0)+(x,y)=(x,y)$. So this means $x_0x=x$ and $y_0y=y$. What do you think these $x_0, y_0$ would be?
Now the negative of a vector $(x,y)$ is a vector $(x',y')$ such that $(x,y)+(x',y')=(x_0,y_0)$. This implies $xx'=x_0$ and $yy'=y_0$. Given the values of $x_0$ and $y_0$ from the previous paragraph, what can you conclude about $x'$ and $y'$.
And the definition of scalar multiplication is clear enough: $\frac13(9,15)=(9^{\frac13},15^{\frac13})$.
If you are stuck somewhere, feel free to ask for more details.
Hope this helps.
Best Answer
For question 1) the answer is "usually no". For example, let $n$ be a positive integer greater than $1$ and let $V$ be a one dimensional subspace of $\mathbb R^n$. The zero vector in $V$ is certainly not the scalar $0$. The reason I say "usually" no is that if you view $\mathbb R$ as a vector space over $\mathbb R$, then the zero vector happens to be equal to the zero scalar. You could cook up some other examples like that.
For question 2, speed is a scalar and velocity is a vector. If an object has speed $0$, then its velocity is the zero vector, but its speed is not equal to its velocity. (They could not be equal because they are not even the same type of mathematical object.)
To be more concrete, let's say that I introduce a coordinate system in my lab and measure that my speed (in meters/sec) is the number $0$. Then my velocity (in meters / sec) is $(0,0,0)$. And $0 \neq (0,0,0)$.
Here's another way to make the same point. Suppose that a particle's position at time $t$ is $f(t)$, where $f:\mathbb (a,b) \to \mathbb R^3$ is a differentiable function. The particle's velocity at time $t_0$ is $f'(t_0)$, and the particle's speed at time $t_0$ is $\| f'(t_0)\|$. Suppose that the particle's speed at time $t_0$ is the number $0$. Then the particle's velocity at time $t_0$ is $f'(t_0) = (0,0,0)$. And again, $0 \neq (0,0,0)$.