First: linear operators don't have bases. Vector spaces (and subspaces) have bases. You'll note that the question does not ask for a basis for the linear operator, it asks for a basis of the range of $\tau$, and for a basis of the nullspace of $\tau$; and it so happens that both of those are vector spaces, so we can talk about bases for them.
Now, $V$ is $4$-dimensional as a vector space over $\mathbb{R}$; a possible basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.
You should also check and make sure that $\tau$ is a linear transformation if you haven't done so.
Let's consider first the nullspace of $\tau$ (it is simpler than the range). When is $(z_1,z_2)$ in the nullspace of $\tau$?
$(z_1,z_2)\in\mathbf{N}(\tau)$ if and only if $\tau(z_1,z_2)=(0,0)$, if and only if $z_1-\overline{z_1}=0$ and $z_2-\overline{z_2}=0$, if and only if $z_1=\overline{z_1}$ and $z_2=\overline{z_2}$, if and only if $z_1$ is real and $z_2$ is real. That is, the nullspace is the span of $(1,0)$ and $(0,1)$, and they give you a basis.
What is the range of $\tau$? Note that if $z = a+bi$ is a complex number, then $z-\overline{z} = (a+bi) - (a-bi) = 2bi$. So another way of writing $\tau$ is:
$$\tau(z_1,z_2) = \Bigl( 2i\mathrm{Im}(z_1), 2i\mathrm{Im}(z_2)\Bigr).$$
So you should notice that the image of $\tau$ is contained in the span of $(i,0)$ and $(0,i)$.
Is the image of $\tau$ exactly equal to $\mathrm{span}((i,0),(0,i))$?
Given the answer to the latter question, can you find the answer to your question of deciding whether $V = \tau(V)\oplus \mathbf{N}(\tau)$?
The basic principal is that elementary row options do not change the rowspace of a matrix. To prove that, you could demonstrate that the processes of scaling a basis element, interchanging two basis elements, and adding multiples of one basis element to another, all result in a new basis.
So, if you perform elementary row operations and change it entirely to row-echelon form, you are left with a matrix that has some zero rows and some nonzero rows. The row-echelon shape makes it obvious that the nonzero rows are linearly independent. Can you see why?
Since this is the case, the remaining rows are a basis for the rowspace of the matrix.
Basic row operations do not change the (right) nullspace either, so you can use the new matrix to work with the nullspace. Using that matrix instead of the original matrix, it will be particularly easy to solve the equation $A'x=0$ with back substitution to discover what is in the nullspace.
Best Answer
The trace function is given by the formula $\DeclareMathOperator\tr{tr}\tr(A) = \sum_{i = 1}^n a_{ii}$, where $A$ is an $n \times n$ matrix $(a_{ij})$. Since the off-diagonal entries don't appear, for computations it is a simplification to think of $A$ as being the sum of two matrices, $A = D + B$, where $D$ is the diagonal part and $B$ is the off-diagonal part; $\tr(A) = \tr(D)$ and $\tr(B) = 0$. This turns the problem into two smaller problems:
Find a basis for the kernel of the map $(x_1, \dots, x_n) \mapsto x_1 + \dots + x_n$, where we denote the diagonal entries of $D$ by $x_i = a_{ii}$.
Find a basis for the set of matrices $B$ whose diagonal is zero.
The first one is amenable to the usual method for getting a basis for the kernel of a matrix; the second one is amenable to direct inspection. Does that help? (I don't think there's any further value in solving the problem here.)