The sigma-algebra generated by $1_{[0,1/2]}$ is simply
$$
\bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}.
$$
It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomain of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)
The situation for $1_{[1/4,3/4]}$ is similar.
The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.
The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.
Check that both sides are equal!
Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?
For Q1 and Q2: Yes, you are right. Actually, every countably generated $\sigma$-algebra is induced by a real valued function, so it comes from a separable pseudometrizable space.
For Q3: Yes, the Borel $\sigma$-algebra on $\mathbb{R}$ is a counterexample. You can add a countable number of points to any topology generating the usual Borel sets without changing the generated $\sigma$-algebra.
For Q4: The $\sigma$-algbera consisting of countable sets and sets with countable complements on $\mathbb{R}$ is a sub-$\sigma$-algebra of the Borel $\sigma$-algebra and is generated by the cofinite topology, which is not second countable. Since the countable-cocountable $\sigma$-algebra is also not countably generated, it doesn't come from any second countable topological space.
Best Answer
Hint: Each $\mathbf 1_{A_n}$ is measurable with respect to $X=\sum\limits_k\frac1{3^k}\mathbf 1_{A_k}$.