After a long night of studying I finally figured out the answer to these. The previous answers on transformation were all good, but I have the outlined steps on how to find $\mathrm{im}(T)$ and $\ker(T)$.
$A = \left(\begin{array}{crc} 1 & 2 & 2 & -5 & 6\\ -1 & -2 & -1 & 1 & -1\\ 4 & 8 & 5 & -8 & 9\\ 3 & 6 & 1 & 5 & -7 \end{array}\right)$
(1) Find $\mathrm{im}(T)$
$\mathrm{im}(T)$ is the same thing as column space or $C(A)$. The first step to getting that is to take the Transpose of $A$.
$ A^T = \left(\begin{array}{crc} 1 & -1 & 4 & 3 \\ 2 & -2 & 8 & 6 \\ 2 & -1 & 5 & 1 \\ -5 & 1 & -8 & 5 \\ 6 & -1 & 9 & -7 \end{array}\right)$
once that's done the next step is to reduce $A^T$ to Reduced Row Echelon Form
$ \mathrm{rref}(A^T) = \left(\begin{array}{crc} 1 & 0 & 1 & -2 \\ 0 & 1 & -3 & -5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$
now on this step I honestly don't know the reasons behind it, but the thext thing you do is take the rows and that's your answer. so that:
$\mathrm{im}(T)\ = \begin{align*} \operatorname{span}\Bigg\{\left(\begin{array}{crc} 1 \\ 0 \\ 1 \\ -2 \end{array}\right), \left(\begin{array}{crc} 0 \\ 1 \\ -3 \\ -5 \end{array}\right)\Bigg\} \end{align*}$
(2) Find $\ker(T)$
$ker(T)$ ends up being the same as the null space of matrix, and we find it by first taking the Reduced Row Echelon Form of A
$ \mathrm{rref}(A) = \left(\begin{array}{crc} 1 & 2 & 0 & 3 & -4\\ 0 & 0 & 1 & -4 & 5\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$
we then use that to solve for the values of $\mathbb R^5$ so that we get
$\begin{align*} \left(\begin{array}{crc} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array}\right) = r\left(\begin{array}{crc} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right) + s\left(\begin{array}{crc} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{array}\right) + t\left(\begin{array}{crc} 4 \\ 0 \\ -5 \\ 0 \\ 1 \end{array}\right) \end{align*}$
from that we arrange the vectors and get our answer the vectors and that gives us our answer
$\begin{align*} ker(T) = span\Bigg\{\left(\begin{array}{crc} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right), \left(\begin{array}{crc} -3 \\ 0 \\ 4 \\ 1 \\ 0 \end{array}\right), \left(\begin{array}{crc} 4 \\ 0 \\ -5 \\ 0 \\ 1 \end{array}\right)\Bigg\} \end{align*}$
and that's that.