Polyhedrons exclusively made out of even sided polygons I know that the cube is the only 3 d shape which falls in polyhedrons but still is composed of squares, exclusively although its a even sided shape.
I have noticed that after square there is no single polyhedron which is exclusively made out of a polygon with even number of sides. For example the hexagon, which does not have a polyhedron made exclusively out of it, although one exists for the pentagon, the dodecahedron.
Why is it so? And is there any example of a even sided polygon, other than square having a polyhedron made exclusively out of it?
 A: I assume you are talking about regular hexagons, etc.
You cannot make a polyhedron out of hexagons, septagons, or any larger regular polygon alone. The reason is because their angles are too big. If you try to fit three hexagons together meeting a vertex, they are forced to lie in the same plane because their three $120$° angles add up to a full $360$°. Going bigger, there is not even enough room for three septagons or octagons to meet at vertex. 
It just happens to be the case that the pentagon is the regular polygon with most sides such that three can meet in three dimensions, and there is only one even number in the range $3,4,5$. 
By the way, there are other polyhedra all of whose faces are squares. Consider gluing $6$ cubes to the faces of a central cube.

Edit: Actually, there is a much better answer to your question. Even if you allow irregular hexagons (or octagons, etc), it is impossible to have a polyhedron whose faces are all hexagons. You can prove every polyhedron has a face with either $3,4$ or $5$ sides as follows. 
If all faces are hexagonal or greater, than $2E \ge 6F$. This follows by the handshaking lemma, applied to the "dual graph" of the polyhedron; every face has six or more sides, so the sum of the degrees, which is equal to $2E$, is at least $6F$. Similarly, every vertex has at least three neighbors, so $2E\ge 3V$. Therefore,
$$
V-E+F = (V-\tfrac23 E) + (F-\tfrac13 E)\le 0 + 0 <2,
$$ 
contradicting Euler's formula, $V-E+F=2$.

2nd Edit: One final thought! The proof above relied on Euler's formula, which is only valid for polyhedra whose surfaces are homeomorphic to a sphere. If one allows for toroidal polyhedra, then it is possible to have only hexagonal faces:

Image source: https://mathematica.stackexchange.com/a/39885/57584
A: Regular polyhedra are the tetrahedron (3 triangles per vertex, 4 triangles in total), the octahedron (4 triangles per vertex, 8 triangles in total), the icosahedron (5 triangles per vertex, 20 triangles in total), the cube (3 squares per vertex, 6 i squares in total), and the dodecahedron (3 pentagons per vertex, 12 pentagons in total). 
Less than 3 faces per vertex does not make a corner with volume, the sum of the face corner angles around a specific vertex clearly has to be less than the full circle in order to be not flat.
If you'd allow for non-convex regulars, then you'd have also the great stellated dodecahedron (3 pentagrams per vertex, 12 pentagrams in total), the small stellated dodecahedron (5 pentagrams per vertex, 12 pentagrams in total), the great icosahedron (5 triangles per vertex running around twice, 20 triangles in total), and the great dodecahedron (5 pentagons per vertex running around twice, 12 pentagons in total).
The loss about allowing for non-convexity is that the Euler equation does not hold in general without adjustment. 
Still keeping convexity, but relaxing the quest that all vertices are to be alike, then you still get more figures with all the same (regular polygonal) faces. In fact you can have convex deltahedra (all faces are triangles, i.e. Delta shaped) with a total of 4, of 6, of 8, of 10, of 12, of 14, of 16, and of 20. (The numbers 4, 8, and 20 occur already above. The others form a subset of the Johnson solids.)
Closer to the title would be the relaxation of alike faces, while keeping those still regular polygonal, and all vertices alike. When aditional restricting to even sided faces only, then there are the truncated octahedron with 1 square and 2 hexagons per vertex, the great rhombicuboctahedron with 1 square 1 hexagon, and 1 octagon per vertex, and the great rhombicosidodecahedron with 1 square, 1 hexagon, and 1 decagon per vertex.
Hope that this small enlisting would provide you some taste of polyhedron research. You might start solving questions about the here missing number of 4 pentagrams per vertex in the non-convex regulars, or the missing number of 18 triangles in total for convex deltahedra. Or, most easily, what would happen with 1 square, 1 hexagon, and 1 dodecagon per vertex (provided all are regular polygons)?
--- rk
