Can a matrix have the same range and nullspace? If you can pick any $3\times 3$ matrix, is there a matrix that its $R(A) = N(A)$?
If you can pick any $4\times 4$ matrix, is $R(A) = N(A)$ possible?
Here, $R(A)$ is the range of matrix A, and $N(A)$ is the nullspace of matrix A
I thought the answers to both questions would be "no" because
$R(A)$ is obtained from $Ax=b$, where $b \not= 0$
$N(A)$ is obtained from $Ax=0$.
The vectors in both spaces will never be the same. I wonder why the question would be structured in such a way.

Your basic misunderstanding is that because the definitions of range and null space involve different equations, they cannot have common vectors. (The question that was posed to you goes further, in requiring them to be identical sets, but you seem to argue that even having some common nonzero vector is impossible.) But the equations are not the whole story; range and null space are linked in a different manner to their equations. The vague "obtained from" in your question takes two different meanings in the two cases.
The null space is simply the solution set for the unknown $x$ in the equation $Ax=0$. But the range is not the solution set for the unknown $x$ in the equation $Ax=b$ (in fact, for what value of$~b$ would that be?), it is the set of vectors $b$ for which this equation has a non-empty solution set. If for some $x$ one sets $b=Ax$, then that $b$ is automatically in the range space; it might also be in the null space if $Ab=0$. This is not directly related to the first equation. In fact the equations combined give $A(Ax)=0$ and for some matrices $A$ it is possible to have this, even without having $x=0$.
Since the question lets you choose $A$ freely, it is really about what contraints there are thet relate the subspaces $R(A)$ and $N(A)$ as $A$ varies. It turns out there is one, and only one, such constraint, namely that $\dim(R(A))+\dim(N(A))=n$ (the size of the matrix), as the rank-nullity theorem says. Then having $R(A)=N(A)$ requires $n$ to be even, so it cannot be done for $n=3$, but it can for $n=4$.
