1)
Define a linear transformation $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with $ker(f)=f(\mathbb{R}^2)$
Solution
At first this was not easy for me to wrap my head around but after some thought it seems simple: Everything that the function outputs is mapped to 0 if it is fed again into the function. So I put:
$f(\begin{bmatrix}r_1\\r_2\end{bmatrix}) = \begin{bmatrix}r_2\\0\end{bmatrix}$ for all $r = \begin{bmatrix}r_1\\r_2\end{bmatrix} \in \mathbb{R}^2$
2)
Define a linear transformation $f$ so that $f \circ f=f$. $f \neq 0$ and $f \neq id_{\mathbb{R}^2}$
Solution
$f(x) = \begin{bmatrix}1\\0\end{bmatrix}$ for all $x \in \mathbb{R}^2$
Because $(f \circ f)(x) = f(f(x)) = f(x) = \begin{bmatrix}1\\0\end{bmatrix}$
Are these solutions and the notation correct?
The first one is correct, the second is not as $f$ is not linear. Remember that linear mappings fulfil $f(0) = 0$ (where $0$ is to be understood as a vector).
Hint for 2): Set up $$ f(x_1, x_2) := \begin{pmatrix} a & b \\ c& d \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. $$ Calculate $$ \begin{pmatrix} a & b \\ c& d \end{pmatrix}^2 = \begin{pmatrix} a^2 + bc & ab + bd \\ ca + cd & cb + d^2 \end{pmatrix} \overset{!}{=} \begin{pmatrix} a & b \\ c & d \end{pmatrix} . $$ Now choose an adequate combination of $a, b, c, d \in \mathbb{R}$.