1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?
On a curve of genus $g$, a general divisor of degree $d \le g-1$ has no sections. Of course, if $d>0$ then it is ample.
$K_X$ on a hyperelliptic curve is globally generated but not very ample.
Look at $L=\mathcal O(1)$ on a plane curve of genus $d$. Then from
$$ 0\to \mathcal O_{\mathbb P^2}(1-d) \to \mathcal O_{\mathbb P^2}(1) \to \mathcal O_C(1)\to 0$$
you see that $H^1(\mathcal O_C(1))=H^2(\mathcal O_{\mathbb P^2}(1-d))$ which is dual to $H^0(\mathcal O_{\mathbb P^2}(d-4))$. So that's nonzero for $\ge4$.
2. Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^k$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?
Again, just look at the divisor of a degree 1 on a curve of genus $g$. You need $k\ge g$, so you see that there is no bound in terms of the dimension.
It turns out that a better right question to ask is about the adjoint line bundles $\omega_X\otimes L^k$ ($K_X+kL$ written additively). Then the basic guiding conjecture is by Fujita, and which says that for $k\ge \dim X+1$ the sheaf is globally generated, and for $k\ge \dim X+2$ it is very ample. This is proved for $\dim X=2$, proved with slightly worse bounds for $\dim X=3$. For higher dimensions the best result is due to Angehrn-Siu who gave a quadratic bound on $k$ instead of linear. There are some small improvements for some special cases.
3. If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X\subset \mathbb P^N$, I can eventually get a finite morphism $X\to \mathbb{P}^d$, where $d$ is the dimension of $X$. But what if I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb P^d$?
But of course $L$ gives a morphism $f$, and it follows that $f$ is finite: $f$ contacts no curve so $f$ is quasifinite, and $f$ is projective (since $X$ was assumed to be projective). And quasifinite + proper = finite.
Answer: no. Example: take a simple abelian surface X with real multiplication by Q($\sqrt{d}$) (where d is a square-free positive number). X has Picard number 2, and the intersection form on N^1(X) diagonalises over Q to diag(a,-b) where b/a=d. The nef cone is just the cone of classes x in N^1(X) with x^2 >= 0 (more precisely, the part of this cone which also satisfies x.h >=0 for a chosen ample class h), and a simple computation shows that the boundary rays of this cone are irrational (by square-freeness of d). Now for any abelian variety the effective cone is equal to the nef effective cone, which equals the union of the ample cone with the rational boundary faces of the nef cone. In our example the nef cone has no rational boundary faces, so the ample cone and the effective cone (minus zero) coincide.
Best Answer
Here's a partial answer. Suppose we're in characteristic 0 (Fujita would be assuming this), and that $rank(E)=2$. By cor 7.6 of Hartshorne's ample vector bundles paper, it suffices to check that $deg(E)>0$ and $deg(L)>0$ for ay quotient line bundle. From Riemann-Roch as in Piotr's comment, we get $$deg(E) + rank(E)(1-g) = h^0(E)\ge 0$$ which implies positivity of $deg(E)$. On a curve $H^1(E)=0$ implies the vanishing for any quotient bundle, and so in particular for $L$. Combing this with the above argument, gives $deg(L)>0$.
I think this can be pushed, but I'd better back to the less fun things that I'm supposed to be doing now.