The isomorphism between Hilbert scheme of n points and moduli space

I've been reading the isomorphism between Hilbert scheme of n points and the moduli space $M(1,0,1-n)$, mainly following Huybrechts's book Lectures on K3 Surfaces, Chapter 10. On Page 215, he stated that let $X$ be a K3 surface and $E\in M(1,0,1-n)$ a coherent sheaf, then $E = I_Z\otimes L$ for some dimension zero closed subscheme $Z$ of $X$ and line bundle $L$ on $X$. Then we have an exact sequence \begin{equation*} 0\rightarrow I_Z\otimes L\rightarrow L\rightarrow \mathcal{O}_Z\rightarrow 0, \end{equation*} which shows that \begin{equation*} (1,0,1-n)=\nu(E) = \nu(I_Z\otimes L) = \nu(L)-\nu(\mathcal{O}_Z) = (1,c_1(L),c_1(L)^2/2+1-h^0(\mathcal{O}_Z)). \end{equation*} Hence $L\cong \mathcal{O}_X$.

Since I am not familiar with the first Chern class, I have the following questions:

1. Why the sequence is exact? We have an exact sequence \begin{equation*} 0\rightarrow I_Z\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_Z\rightarrow 0, \end{equation*} and since $L$ is locally free, tensoring with $L$ preserves the exactness. But why $\mathcal{O}_Z\otimes L \cong \mathcal{O}_Z$?

2. What's the definition of the first Chern class? It seems that there are two possible definitions: the first one is to use the correspondance between line bundles and divisors, then for any line bundle we have a codimension one cocycle in $CH^1(X)$ and there is a correspondance between $CH^1(X)$ and $H^2(X,\mathbb{Z})$; the second one is to use the fact that every algebraic K3 surface is a projective complex K3 surface.

3. Why are $c_1(\mathcal{O}_X)$ and $c_1(i_*(\mathcal{O}_Z))$ zero, here $i:Z\rightarrow X$ is the inclusion? If we use the first definition, then $\mathcal{O}_X$ corresponds to the zero divisor and hence its first Chern class is zero. But I don't know how to compute $c_1(i_*\mathcal{O}_Z)$.

Then on Page 216, proposition 3.8, he stated that the isomorphism $\operatorname{Hilb}_{X/k}^n\rightarrow M(1,0,1-n)$ is given by $Z\mapsto I_Z$. My fourth question is why $I_Z$ is semistable? I guess it may be a line bundle, and thus is stable. Or somehow follows the exact sequence $0\rightarrow I_Z\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_Z\rightarrow 0$ and $\mathcal{O}_Z$ has dimension zero and hence $I_Z$ is somehow similar to $\mathcal{O}_X$.

It seems that there is a similar question here. But it didn't answer my question. So I ask again. Thanks for any help.