In general, you want to use Ramsey's theorem when you want to construct a large/infinite set of objects such that any $n$-sized group of them has a certain property. With that in mind, here's an example of how you can use Ramsey's Theorem on your two problems:
For (1), draw a blue edge between two elements of the sequence if they are ascending (i.e. the one that comes later in the sequence is the larger of the two) and draw a red edge if they are descending. By Ramsey's Theorem, we are done.
(2) is harder. The first obstacle is that there's no obvious disjunction: there are pairs of circles which do not touch yet are closer than $M$ apart, which really throws a wrench into the "obvious" Ramsey approach. However, we can try naively coloring any such pair of circles a third color and working with the 3-color version of Ramsey.
There's another obstacle, though. If we somehow get a set of $k$ circles such that each pair has a common point, there's still no guarantee that all of the circles have a common point. We can fix this too: instead of just saying that pairs of circles should have a common point, say that their centers should be less than 1/1000 units apart. If in a set of $k$ circles, any pair satisfies this, then all $k$ circles definitely have a point in common.
With this in mind, color a pair of circles red if their centers are less than 1/1000 units apart, blue if they're more than $M$ units apart, and orange if neither. Now, note that there's a limit on the number of circles we can have in an all-orange subgraph. In particular, any all-orange subgraph must fit inside a huge circle of radius $M$ - otherwise, some pairs will be blue. Also, each circle in an all-orange subgraph must have a small circle of radius 1/1000 to itself - otherwise, some pairs wil be red.
Therefore, by Pigeonhole, there's some number $B$ such that no all-orange subgraph has size $B$ or greater. Now we can finish it off: by Ramsey's Theorem, there's a number of circles so large that there's either a $B$-sized all-orange subgraph, a $k$-sized all-red subgraph, or a $k$-sized all-blue subgraph. The $B$-sized all-orange subgraph is impossible, so it must be one of the other two possibilities, and we are done.
Note that (2) might also be solved just using the Pigeonhole Principle. Ramsey's Theorem is almost never your only recourse, although it's really helpful to know!