Is these representations completely reducible?
Definition:
A linear representation is said to be completely reducible if every invariant subspace has an invariant complement.
But I have no idea how to apply the definition, and I know that the answer for 2 is no while that for 3 is yes, could you please clarify for me how to check the definition in each case?
Consider polynomials of degree less than or equal to 1 $$ V_1=\mathbb C\cdot x\oplus\mathbb C\cdot 1, $$ this is clearly invariant under $F(t)$ as $F(t)$ does not change degree. Similarly consider $$ V_0=\mathbb C\cdot 1, $$ the space of constant functions. It is clear that $V_0\subset V_1$ and that they are both invariant under $F$. However, if we consider the complement of $V_1$ in $V_0$, $W$, then there exists some $a\in\mathbb C$ such that $x+a\in W$. If $W$ is invariant, then we have that $L(1-a)(x+a)=x+1\in W$. $W$ is a vector space so $x+a-L(1-a)(x+a)=1\in W$. This means $V_0\subset W$, a contradiction. Hence the space is not completely reducible.
$\alpha\in$ End$(V)$, with $V\cong\mathbb C^n$, so $\alpha$ has Jordan Normal form, but its characteristic polynomial has no multiplicities, so it is diagonalisable. In the diagonal basis we have that $\alpha=\text{diag}(\lambda_1,\dots,\lambda_n)$. So $$ F(t)=\text{diag}(e^{t\lambda_1},\dots,e^{t\lambda_n}), $$ and $$ V=\mathbb C_{\lambda_1}\oplus\dots\oplus\mathbb C_{\lambda_n}, $$ where $\mathbb C_{\lambda_i}$ is the one dimensional representation $v\mapsto e^{t\lambda_i}v$. So $V$ is completely reducible.