Uniform, strong and weak convergence of the operators

First of I just want to provide the definitions I have for the three concepts. Let $A_n: X \to Y$.

$A_n$is said to converge to $A:(n\to \infty)$

$$\text{Uniformly : }\|A_n-A\|=0 \\ \text{Strongly : }\lim_{n \to \infty}A_n(x) = A(x) \\ \text{Weakly :} \lim_{n \to \infty}y^*(A_nx)=y^*(Ax) :\forall y^*\in X^*$$

Sometimes, I come across that strong convergence is $\lim_{n \to \infty}\|A_n(x)-A(x)\|=0$, which is logical.

Anyway: I am to question the convergence of the following:

$$a.) \ \ D_n : C[0,1]\to C[0,1]: D_n(x(t))=t^n(1-t)x(t),x \in C[0,1] \\ b.) \ \ E_n : L_2[0,1]\to L_2[0,1]: E_n(x(t))=\int_{0}^{1}t^nsx(s)ds,x \in L_2[0,1] \\ c.) \ \ F_n : \ell_2 \to \ell_2: F_n(x)=(\frac{x_1}{n^2},\frac{x_2}{n^2},\frac{x_3}{n^2},...),x=(x_1,x_2,...) \in \ell_2 \\$$

I also have the theorem that:

Let $\{A_n\}_{n=1}^{\infty}$ be a sequence $A_n:X \to Y$ ($X,Y$ are Banach spaces) such that $\|A_n\|\leq M, M>0.$ Let $E$ be a fundamental set in $X$.

If $\exists \lim_{n \to \infty}A_n(x), \forall x \in E\implies \exists > \lim_{n \to \infty}A_n(x),\forall x \in X$

* ------My solutions-------- *

I know that the fundamental set in $C[0,1]$ is the set $\{1,t,t^2,...\}$ and in $\ell_2: \{(1,0,..),(0,1,0,0,..),(0,0,1,0,..),...\}.$ How about in $L_2[0,1]?$ Do I have to take this route of the theorem. Because, for example for $a.$ I tried:

$\lim_{n \to \infty}A_n(x(t))=0;$(because $t \in (0,1)$) to I check the uniform by:

$$0 \leq \|A_n(x(t))\|_{C[0,1]} \leq \|x\| \lim_{n\to \infty}\max_{t\in[0,1]}(t^n(1-t)) \implies \\ 0 \leq \frac{\|A_n(x(t))\|_{C[0,1]}}{\|x\|} \leq \lim_{n\to \infty}\max_{t\in[0,1]}(t^n(1-t))=0 \implies$$ that this sequence converges uniformly to 0, also strongly and weakly because of inclusion relation of the three.

For the second: $\lim_{n\to \infty}A_n(x(t))= \begin{cases} 0 & \text{if$t \in [0,1)$} \\ \int_{0}^{1}sx(s)ds & \text{if$t = 1$.} \end{cases}$. I don't know how to work with this.

For the third I let the limit go to $\infty$ and question the convergence of the sequence to $(0,0,0,...)$. Is this valid what I am doing in your opinion? There is always the possibility that the sequences don't converge, like having a fixed distance in between two elements, but, what do you all think?

I have my exam in two days, I need to clear this up. Thanks for all input.

The most important thing is to rewrite your sequences so that it is clear what is happening. For example once you see that $F_n$ is just $\mathbb1/n^2$ it is obvious that $F_n\to0$ uniformly.
What does $E_n$ do? $$E_n(x)\,(t)=t^n\int_0^1 s x(s)\,ds = t^n\,\langle \mathrm{id},x\rangle\tag{1}$$ So $$\|E_n(x)\|_2=\|t^n\|_2\,|\langle \mathrm{id},x\rangle≤\|t^n\|_2\,\|x\|_2\,\|\mathrm{id}\|_2=\frac{\|x\|_2}{2(n+1)}$$ and $\frac{\|E_n(x)\|}{\|x\|}≤\frac1{2(n+1)}$, thus $\|E_n\|=\sup_{x\neq0}\frac{\|E_n(x)\|}{\|x\|}≤\frac{1}{2(n+1)}$ and $E_n\to0$ uniformly as well.
The way to see it immediatly is to understand that equation $(1)$ is of the form $E_n= A_n \circ B$, with $B:L^2\to\Bbb C, x\mapsto \langle\mathrm{id},x\rangle$ and $A_n:\Bbb C\to L^2, z\mapsto (t\mapsto t^n\,z)$. It follows $\|E_n\|≤\|A_n\|\,\|B\|$, where $\|A_n\|\to0$.
You have seen correctly that $D_n$ converges uniformly also.