On
If they don't need to be regular, then you can have any number of polygons meeting at a vertex. In this case, the angle at a given vertex can be as small as one wants to, and therefore there can be as many meeting at a given vertex as one desires.
If they have to be regular, but they need not be different from each other, I guess the maximum is 6. The smallest angle between two adjoining sides in a regular polygon is 60 (in a triangle). Therefore you can fit 6 in 360º.
If they have to be regular and different from each other, I think the maximum is 3. You can use a triangle (60º) a square (90º) and a pentagon (108º), and there will be 102º left, which is smaller than needed to fit an hexagon (120º).
On
The answer strongly depends on what are the rules to follow. What is your intended setup?
Shall your derived polyhedron be convex, or does it not need to? Or would you truely intend to speek of polyhedra, or would euclidean or even hyperbolic tilings be allowed as well? Are those "types of polygons" intended to be distinguished by shape, or maybe only by colour?
--- rk
If you are in the Euclidean plane, start with a circle and divide it into $n$ equal pieces using radii from the centre. Each piece can be completed as a polygon with as many additional sides as you choose, approximating the circumference of the circle.
Obviously these are not regular. A square has a right-angle at a vertex, and if you have five different regular polygons meeting at a vertex, four of them will have four sides or more, so will not fit as the total is already more than four right-angles.
In fact you can't fit four different regular polygons either - a triangle and a hexagon make straight line and a square plus a pentagon are more than a straight line - so the four smallest angles don't do it.