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.
Best Answer
The bundles with no derived global sections (more generally the objects $F$ of the derived category $D^b(coh X)$ such that $Ext^\bullet(O_X,F) = 0$) form the left orthogonal complement to the structure sheaf $O_X$. It is denoted $O_X^\perp$. This is quite an interesting subcategory of the derived category.
For example, if $O_X$ itself has no higher cohomology (i.e. it is exceptional) then there is a semiorthogonal decomposition $D^b(coh X) =< O_X^\perp, O_X >$. Then every object can be split into components with respect to this decomposition and so many questions about $D^b(coh X)$ can be reduced to $O_X^\perp$ which is smaller. Further, if you have an object $E$ in $O_X^\perp$ which has no higher self-exts (like $O(-1)$ on $P^2$), you can continue simplifying your category --- considering a semiorthogonal decomposition $O_X^\perp = < E^\perp, E >$. For example if $X = P^2$ and $E = O(-1)$ then $E^\perp$ is generated by $O(-2)$, so there is a semiorthogonal decomposition $D^b(coh P^2) = < O_X(-2), O_X(-1), O_X >$ also known as a full exceptional collection on $P^2$. It allows a reduction of many problems about $D^b(coh P^2)$ to linear algebra.
Another interesting question is when $O_X$ is spherical (i.e. its cohomology algebra is isomorphic to the cohomology of a topological sphere). This holds for example for K3 surfaces. Then there is a so called spherical twist functor for which $O_X^\perp$ is the fixed subcategory.
Thus, as you see, the importance of the category $O_X^\perp$ depends on the properties of the sheaf $O_X$.