According to PlanetMath, a necessary and sufficient condition for a quadrilateral to be cyclic, is that the sum of a pair of opposite angles be equal to 180 degrees.

Concerning part 1, let F be a point on m left of C. I believe there is a theorem that angle FCA = angle ABC (if you need help finding it, let us know). We have angle ADE = 180 - CDE = 180 - FCA. Thus, ADE + ABE = 180. Similarly, it's easy to show that DAB + DEB = 180.