Is the center of the universal enveloping algebra generated by the center of the lie algebra?

If the center of $\mathfrak{g}$ is 0, this only means that $U(\mathfrak{g})$ has no central elements of degree 1, but there could be central elements of higher degree. There is a natural $\mathfrak{g}$-module map $$\mathfrak{g} \otimes \mathfrak{g} \to U(\mathfrak{g}).$$ Now if, say, $\mathfrak{g}$ is semisimple so that it has a nondegenerate Killing form (a bilinear symmetric invariant form), then there are $\mathfrak{g}$-module isomorphisms $$\mathfrak{g} \otimes \mathfrak{g} \cong \mathfrak{g} \otimes \mathfrak{g}^* \cong \text{End}(\mathfrak{g})$$ Clearly $\text{End}(\mathfrak{g})$ has nonzero central elements, namely scalars, and if you take the corresponding element $\sum e_i \otimes f_i \in \mathfrak{g} \otimes \mathfrak{g}$ and its image $\sum e_i f_i \in U(\mathfrak{g})$, you get a central element of $U(\mathfrak{g})$ called the Casimir element.

(We are not quite done; we still have to show that the element we've constructed is $\ne 0$. This follows from the Poincare-Birkhoff-Witt theorem.)

No, the universal enveloping algebra can have a nontrivial center even if the Lie algebra itself has only $0$ as center. The (or at least a) concept you want to look up is "Casimir element; see for example http://en.wikipedia.org/wiki/Casimir_element .

No, it's bigger in general. This is already the case for $\mathfrak{sl}_2$. More generally see Harish-Chandra isomorphism.