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.