In cases A and B, you can find the matrix of the linear transformation with respect to the canonical bases; in case A it is
$$
A=\begin{bmatrix}
2 & -1 \\
-8 & 4
\end{bmatrix}
$$
and in case B it is
$$
B=\begin{bmatrix}
4 & 1 & -2 & -3 \\
2 & 1 & 1 & -4 \\
6 & 0 & -9 & 9
\end{bmatrix}
$$
In general, when you have a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^m$ and $\{e_1,e_2,\dots,e_n\}$ is the canonical basis of $\mathbb{R}^n$, you just write down (as columns), the vectors $T(e_1), T(e_2), \dots, T(e_n)$. A basis for the range can easily be computed by Gaussian elimination.
For case C, you don't have a "canonical basis", but you still can compute the matrix associated to the bases $\{1,x,x^2\}$ of $P_2$ (assuming it's the space of polynomials having degree at most 2) and $\{1,x,x^2,x^3\}$ of $P_3$. Since $T(1)=x=0\cdot1+1x+0x^2+0x^3$, $T(x)=x^2$, $T(x^2)=x^3$, the matrix is
$$
C=\begin{bmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
The rank of this matrix is? And what can you conclude from this?
You are on the right track. With $a_2 = a_1 = a_0 = 0$, you have shown that the kernel of $g$ is the set of all polynomials of degree 2 or less with coefficients equal to zero. That is,
$$\ker(g) = \{0\}.$$
There is a very special basis for this space - $\{0\}$ - and we define the dimension of this space to be zero.
Now, using the rank-nullity theorem (or dimension theorem or whatever else your book or notes may call it), it follows that the dimension of the range of $g$ (or rank of $g$) is $3$. Why? This follows from the dimension of the domain being 3 and the dimension of the kernel of $g$ being $0$.
Since it is clear that no quadratic coefficient is present in the range, it follows that the range of $g$ is
$$\{ b_3x^3 + b_1x + b_0 | b_i \in \mathbb{F} \}$$
where $\mathbb{F}$ is whatever field that you are given here (the real numbers, the complex numbers, the rational numbers, etc.).
At this point, it is easy to see that $\{1, x, x^3\}$ is a basis for the range of $g$.
Best Answer
Choose a basis $1,t,t^2$ for $P_2$, and $1,t,t^2,t^3$ for $P_3$. The the matrix for $T$ is given by:
$$ \tau = \left[\begin{array}{rrr} -1 & 2 & 0 \\ 0 & -1 & -1 \\ -2 & 3 & -1 \\ 1 & -1 & 1 \end{array}\right].$$ It is fairly easy to see that $\tau (2,1,-1)^T= 0$ (ie, the second column is equal to the third column minus twice the first column), and that the first and last columns are linearly independent. Thus these two columns span the range space of $T$. Thus $T(1), T(t^2)$ form a basis for the range.