1) $c_1(X):=c_1(TX)=c_1(\wedge^n(TX))=-c_1(K_X)$ where $TX$ is the holomorphic tangent bundle as in Georges's comment.
2) If $K_X$ is ample, then $c_1(K_X)>0$ in the sense that $c_1(K_X).C>0$ for all irreducible holomorphic curve $C$ in $X$. This is because some positive multiple of $K_X$ is very ample, hence positive.
Let's say $X$ is a smooth projective variety for simplicity.
The short, maybe unhelpful answer is this: the first Chern class of $L_D$ is exactly the element $[D]$ of the Picard group $\operatorname{Pic}(X)$ canonically defined by the divisor $D$.
To say something more informative, one has to say what one wants to compute $c_1(L_D)$ in terms of. In your example, $\operatorname{Pic}(\mathbf P^3)$ is isomorphic to $\mathbf Z$, with the isomorphism taking a divisor to its degree. (That isn't 100% obvious: there is a detailed proof in Shafarevich.) So the Picard group is generated by $[H]$, the class of a plane.
Your $D$ is a cubic (degree 3) hypersurface in $\mathbf P^3$, so its class in the Picard group is $[D]=3[H]$. (If you want to phrase this in terms of the line bundle, you can say $c_1(L_D)=3[H]$.)
For a general $X$, the situation will be more complicated. The Picard group $\operatorname{Pic}(X)$ might not be generated by a single element, or it might even not be finitely generated. (Example: an elliptic curve.) One way to deal with this is to work in the Néron–Severi group $NS(X)$ instead. Roughly speaking, $NS(X)$ only remembers the numerical properties of a divisor class or line bundle — in other words, its intersection numbers with curves in $X$. The great advantage is that it is always finitely generated (the so-called Theorem of the Base.) Also there is a quotient map $\operatorname{Pic}(X) \rightarrow NS(X)$, so every divisor or line bundle has a well-defined class in $NS(X)$ too.
Now suppose you have fixed a basis of $NS(X)$. To calculate the class of $D$ or $L_D$ in $NS(X)$ in terms of that basis, one calculates the intersection numbers $D \cdot C$ for curves $C$ in $X$. That leads to a system of linear equations for the coefficients of $D$ in the chosen basis, and solving the system tells you the (numerical) class of $D$ in terms of the chosen basis.
Best Answer
If $M$ is a complex manifold of dimension $n$, the exact sequence of sheaves $$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$ yields a morphism of groups in cohomology (part of a long exact sequence) $$c_1:H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)$$
Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the morphism $$c_1:Pic(M)\to H^2(M,\mathbb Z)$$
This morphism coincides with the first Chern class defined (for $C^\infty$line bundles) in differential geometry in terms of curvature of a connection.
If now $M$ is a compact Riemann surface (= of dimension $1$), we may evaluate a cohomology class $c\in H^2(M,\mathbb Z) $ on the fundamental class$[M]$ of the Riemann surface $M$, thereby obtaining an isomorphism $I: H^2(M,\mathbb Z) \xrightarrow {\cong} \mathbb Z: c\mapsto \langle c,[M] \rangle$, which composed with the first Chern class yields the degree for line bundles: $$deg=I\circ c_1:Pic(M)\to \mathbb Z: L \mapsto deg(L)=\langle c_1(L),[M] \rangle$$
Finally, if $L$ is associated to the divisor$D$ i.e. $ L=\mathcal O(D)$, we have the pleasantly down to earth (but highly non tautological!) formula $$deg(\mathcal O(D))=deg (D)$$
which says that the degree on the left, obtained by sophisticated differential geometry, can be computed in a childishly simply way by computing the degree of the divisor $D\in \mathbb Z^{(X)}$ seen purely formally as an element of the free abelian group on $X$.
All this is explained in Griffiths-Harris, Principles of algebraic geometry, Chapter 1.