After a long night of studying and a bit of help I was able to understand the exercise and want to provide a guide to it and similar exercises.
$L = [v^1,v^2,v^3,v^4,v^5]$
$v^1 = \begin{pmatrix} -1 \\3\\1 \end{pmatrix}, v^2 = \begin{pmatrix} 2 \\-6\\-2, \end{pmatrix}, v^3 = \begin{pmatrix} 5 \\1\\-3, \end{pmatrix}, v^4 = \begin{pmatrix} 4 \\4\\-2, \end{pmatrix}, v^5 = \begin{pmatrix} -11 \\1\\7, \end{pmatrix}$
$\Psi_L: \mathbb{K^t} \rightarrow V, (\zeta_1, ..., \zeta_t)' \mapsto \sum^t_{i=1} \zeta_iv^i$ (this is putting a vector in as a skalar for $v^1,...,v^5$)
(a) Calculate a Basis of the Kernel($\Psi_L$)
(b) Calculate a Basis of the Image($\Psi_L$)
1. Rendering the 5 vectors you can see, that $v^1 = v^2$, which results because $v^1 \cdot (-2) = v^2$.
2. All vectors are in the same plane $\Rightarrow$ there are $2$ Vectors so that one can write as a $lin(v^x,v^y)$
(a) Calculate Kernel($\Psi_L$) Now, we should first clarify what we are looking for. When mapping an $\mathbb{R^5}$ Inputvector with the linear image, we get a $\mathbb{R^3}$ outputvector.
With the kernel we are searching for those $\mathbb{R^5}$ vectors, that map onto $\begin{pmatrix} 0 \\0\\ 0\end{pmatrix}$.
Now for example $\begin{pmatrix} 3 \\2\\ -1\\7\\0 \end{pmatrix}$ is mapped as follows:
$3\begin{pmatrix} -1 \\3\\1 \end{pmatrix} +2\begin{pmatrix} 2 \\-6\\-2, \end{pmatrix} -1\begin{pmatrix} 5 \\1\\-3 \end{pmatrix} +7\begin{pmatrix} 4 \\4\\-2, \end{pmatrix} +0\begin{pmatrix} -11 \\1\\7, \end{pmatrix} = \begin{pmatrix} 24\\24\\-12, \end{pmatrix} $
This is exactly the same as: $\begin{pmatrix} -1&2&5&4&-11 \\3&-6&1&4&1\\1&-2&-3&-2&7 \end{pmatrix} \begin{pmatrix} 3 \\2\\-1\\7\\0 \end{pmatrix} $
So $\begin{pmatrix} -1&2&5&4&-11 \\3&-6&1&4&1\\1&-2&-3&-2&7 \end{pmatrix}$ is $\Psi$ and we want to know, which vectors of $\Psi$ map onto $\begin{pmatrix} 0 \\0\\ 0\end{pmatrix}$
Now to do this we of course calculate $\begin{pmatrix} -1&2&5&4&-11&0 \\3&-6&1&4&1&0\\1&-2&-3&-2&7&0 \end{pmatrix}$, which is $\begin{pmatrix} 1&-2&0&1&1&0 \\0&0&1&1&-2&0\\0&0&0&0&0&0 \end{pmatrix} \Rightarrow \mathbb{L_{A,0}} = \begin{pmatrix} 0\\0\\0\\0\\0 \end{pmatrix} + lin\begin{pmatrix} \begin{pmatrix} 2\\1\\0\\0\\0 \end{pmatrix}, \begin{pmatrix} -1\\-1\\0\\1\\0 \end{pmatrix}, \begin{pmatrix} -1\\2\\0\\0\\1 \end{pmatrix}\end{pmatrix}$
So if we were to put any of vector of this linearcombination in $\Psi$ we would map onto the zero-vector.
(b) Now, calculate the basis of the Image($\Psi_L$)
So first of, what is that supposed to mean? $\Psi$ maps a vector $u \in \mathbb{R^5}$ into $\mathbb{R^3}$. Now we want to look at $\mathbb{R^3}$ and dertermine, how the space of the image looks like. We could have a point, a line, a plane or a space. To find out what it is, we have to be aware that each vector is made of $3$ coordinates: $\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}$ So we put $v^1,...,v^5$ into a matrix, with the first column for $x^1$, 2nd for $x^2$ and 3rd for $x^3$. Putting it in a matrix lets us dissolve those vectors and give us only those, that are really needed to span the "image-space":
$\begin{pmatrix} -1&3&1 \\2&-6&-2\\5&1&-3\\ 4&4&-2\\-11&1&7\end{pmatrix} \leadsto \begin{pmatrix} 1&0&\frac{5}{8} \\0&1&\frac{-1}{8}\\0&0&0\\ 0&0&0\\0&0&0\end{pmatrix}$
Now remember, the solution out of this are vectors (must be 2, as we can see in the rendering), and therefore we have to read the lines and transpose them back into vectors: $s^1 = \begin{pmatrix} 1 \\0\\ \frac{-5}{8}\end{pmatrix}, s^2 = \begin{pmatrix} 0 \\1\\ \frac{1}{8}\end{pmatrix}$
Now, $s^2$ and $s^2$ should, geometrically span the same plane as we saw in the image above, that was spanned by $v^1,...,v^2$, and as we can see in the image below, they do:
