$v \times w$ is a bilinear map, antisymmetic and $u \times w =0 \Leftrightarrow $ collinear in tensor product 
This is my Attempt for part (b):
Let's define: $$\Phi: \mathbb{R}^2 \times \mathbb{R}^2 \longrightarrow \mathbb{R}^2 \otimes \mathbb{R}^2 $$ by the following action: $$\Phi(v \times w) = v \otimes w - w \otimes v$$
Clearly, $\Phi$ is bilinear, since:
$$\Phi((r_1v_1+r_2v_2) \times w) = (r_1v_1+r_2v_2) \otimes w - w \otimes (r_1v_1+r_2v_2) = r_1(v_1\otimes w - w \otimes v_1) + r_2(v_2\otimes w - w \otimes v_2) = r_1\Phi(v_1\times w) + r_2\Phi(v_2 \times w)$$
The same procedure can be applied to obtain: $$\Phi(v \times (r_1w_1+r_2w_2)) = r_1\Phi(v\times w_1) + r_2\Phi(v \times w_2)$$
To prove antisymmetry, Im not sure if the following argument do the right proof. I have that if  $\Phi(v \times w) = v \otimes w - w \otimes v$ and $\Phi(w \times v) = w \otimes v - v \otimes w,$ then we must have that: $$  \Phi(v \times w)= -\Phi(v \times w)$$
Now, for the last statement: $u \times w =0 \Leftrightarrow  $ collinear. I have:
"$\Leftarrow$" Let $<e_1,e_2>$ be a basis for $\mathbb{R}^2$, then if $v,w$ are collinear, we have that: $$v = r_1e_1+r_2e_2$$ $$w = s_1e_1+s_2e_2$$Then:
$$v\times w = (r_1e_1+r_2e_2) \otimes (s_1e_1+s_2e_2)$$Then, using the tensor rules, I manipulate such expression to obtain: $$u\times w = r_1s_1(e_1\otimes(e_1-e_1))+ r_1s_1(e_1\otimes(e_2-e_2)) + r_2s_2(e_2\otimes(e_1-e_1)) + r_2s_2(e_2\otimes(e_2-e_2)) = 0$$
"$\Rightarrow$" I have to prove that if $v \times w =0 \Rightarrow  v,w$ collinear, so I have: $$v \times w =0 \Rightarrow v \otimes w = w \otimes v$$ So by part (a), there exist some $\lambda \in \mathbb{R}$, such that $w =\lambda v$. In other words, is a multiple of $v$, so there are collinear.
My question are:
1) How to prove antisymmetry?
2) Is this the right approach?
3) What is the product that is similar used in Calculus III? 
 A: Firstly, some notation: the operation symbol $\times$ is primarily used for the cartesian product of sets / vector spaces, and the elements of $A\times B$ are most often written as simple pairs $(a,b)$ of elements of $A$ and $B$, and not like $a\times b$.
It is a bit unlucky that the operation defined in the exercise is also denoted by $\times$, and it can be also misleading that the usual notation for tensor product is $a\otimes b$ for an element of $A\otimes B$. 
To clear it up, you meant $\Phi(v,w):=$ what they defined as $v\times w$, i.e. $=v\otimes w\,-\,w\otimes v$.


*

*Antisymmetry is immediate from the definition, compare
$$\Phi(v,w)=v\otimes w\ -\ w\otimes v \\
\Phi(w,v)=w\otimes v\ -\ v\otimes w\,.$$
(Note that antisymmetry also follows from bilinearity and $\Phi(v,v)=0$ for all $v$.)

*Mostly yes. For the collinearity part, you overcomplicated the easy part: $v,w$ are collinear iff $w=\lambda v$ for some $\lambda\in\Bbb R$ or $v=0$. 
So suppose $w=\lambda v$, then $\Phi(v,w)=v\otimes(\lambda v)-(\lambda v)\otimes v=\lambda\,v\otimes v - \lambda\,v\otimes v=0$. 
The other direction is just exercise a) as you noted.

*The cross product of vectors in $\Bbb R^3$ has these properties.

