Jordan canonical form- minimal and characteristic polynomial

A linear operator $T$ on a complex vector space $V$ has a characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^3(x-5)$. Choose all correct options:

1) The Jordan canonical form of $T$ is uniquely determined by the given information.

2) There are exactly $2$ Jordan blocks in the Jordan decomposition of $T$.

3) The operator induced by $T$ on the quotient space $V/ \ \text{ker}(T-5I)$ is nilpotent, where $I$ is the identity operator.

4) The operator induced by $T$ on the quotient space $V/ \ \text{ker}(T)$ is a scalar multiple of the identity operator.

How can you argue that option 3 and 4 are incorrect?

We can see that $V$ has dimension 5. Looking at the minimal polynomial of $T$, we find the Jordan canonical form for $T$ must be $$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{pmatrix}$$ which implies (1) is true and (2) is false.
Since $V = \text{ker}(T^3) \bigoplus \text{ker}(T-5I)$, we have $V/ \ \text{ker}(T-5I)$ naturally isomorphic to $\text{ker}(T^3)$. Thus the induced operator on the quotient space is nilpotent, so (3) is true.
However, if we look at $W = V/ \ \text{ker}(T)$, the induced operator $U$ on $W$ has $x^2(x-5)$ as its minimal polynomial which implies (4) is false. We can see this by considering the Jordan form of $U$. (It's the lower right $4\times4$ submatrix of the above matrix.) Alternatively, let $x\in\text{ker}(T^3)$, $x\notin \text{ker}(T^2)$. Then $x,Tx\notin \text{ker}(T)$. If we let $y\in\text{ker}(T-5I)$, then $$U(x + y + \text{ker}(T)) = Tx + 5y + \text{ker}(T).$$ As $Tx \neq 5x$, we again have that (4) is false.