We have tha matrix \begin{equation*}A:=\begin{pmatrix}1 & 5 & 8 & 2 \\ 2 & 4 & 6 & 0 \\ 3 & 3 & 8 & 2 \\ 4 & 2 & 6 & 0 \\ 5 & 1 & 8 & 2\end{pmatrix}\in \mathbb{R}^{5\times 4}\end{equation*}
Determine a basis $B$ of $\mathbb{R}^4$ and a basis $C$ of $\mathbb{R}^5$ such that $M_C^B(A)$ contains at the left upper corner the unit matrix and everywhere else zeroes.
Could you give me a hint for that?
So do we have to write each column of $A$ as a linear combination of $C$ and the corresponding coefficients have to satisfy the desired form that $M$ will have?
If I understand you correctly, then yes. More explicitly, if $B=\{b_1,\dots,b_4\}$ and $C=\{c_1,\dots, c_5\}$ are bases of of $\mathbb R^4$ and $\mathbb R^5$, respectively, and $M^B_C(A)$ looks like
$$M^B_C(A)=\begin{pmatrix}\alpha_1&\beta_1&\gamma_1&\delta_1\\ \alpha_2&\beta_2&\gamma_2&\delta_2\\ \alpha_3&\beta_3&\gamma_3&\delta_3\\ \alpha_4&\beta_4&\gamma_4&\delta_4\\ \alpha_5&\beta_5&\gamma_5&\delta_5\end{pmatrix},$$
then that means that $b_1$ is mapped to $\alpha_1 c_1+\dots+\alpha_5c_5$, while $b_2$ is mapped to $\beta_1c_1+\dots+\beta_5c_5$, and so on. Now you want it to look like
$$M_C^B(A)=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}.$$
So $b_1$ should be mapped to $c_1$, $b_2$ to $c_2$, and so on. The easiest way to find such bases would be to just take the standard base as $B$. Then look at the vector to which $b_i$ is mapped and choose that as $c_i$. Then choose some additional vector as $c_5$ which completes the base.