What is considered 'identical' for these questions? If mirror-image shapes are allowed, these are all pretty trivial. If not, the problems are rather more difficult...

1. Connect the center to every second vertex.

2. Connect the center to the midpoint of each side.

3. This is the hard one. If you allow mirror images, it's trivial: bisect the hexagon from vertex to vertex, then bisect with a perpendicular to that, from midpoint of side to midpoint of side.

4. This one's neat. Let the side length of the hexagon be 2 (WLOG). We can easily partition the hexagon into equilateral triangles with side 2 (6 of them), which can in turn be quartered into equilateral triangles with side 1. Thus, our original hexagon is partitioned into 24 unit equilateral triangles. Take the trapezoid formed by 3 of these little triangles. Place one such trapezoid on the inside of each face of the original hexagon, so that the long side of the trapezoid coincides with the side of the hexagon. This uses 6 trapezoids, and leaves a unit hexagon in the center as yet uncovered. Cover this little hexagon with two of the trapezoids. Voila. An 8-identical-trapezoid partition.

5. Easy. Do the rhombus partition in #1. Quarter each rhombus by connecting midpoints of opposite sides. This produces 12 small rhombi, each of which is equivalent to two adjacent small triangles as in #4.

Except for #3, all of these partitions can be achieved by breaking up the hexagon into unit equilateral triangles, and then building these into the shapes desired. For #3, though, this would require (since there are 24 small triangles) trapezoids formed from 6 triangles each. The only trapezoid that can be built from 6 identical triangles is a parallelogram; I assume that the poster wouldn't have asked for a trapezoid if you could do it with a special case of trapezoid. At any rate, that parallelogram doesn't work.

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