a few remarks:

the fact that ab = (m^{d})(n^{d}) = (mn)^{d}relies on the fact that multiplication modulo p is commutative.

claim 2) and claim 3) of part 1) are superfluous, Z_{p}* is finite, so closure suffices.

for part 2) claim 1):

i think you want to use a specific a, here; namely, let m be a generator for Z_{p}*, so that |m| = p-1. then it follows from a result on cyclic groups that:

a = m^{d}has order (p-1)/d. thus <a>, which is a subgroup of H, has order (p-1)/d, so (p-1)/d divides |H|.

in particular, i found this statement:

"....m^{d|H|}=1, Therefore, p-1 divides d|H|..."

hard to follow, unless we know something about m (it is trivially true that for any group G, g^{|G|}= e, but that reveals very little about the order of g). i think we need to know that |m| = p-1, for this to tell us that p-1 divides d|H|.

claim 2): again, i think you need to pick an a in H, with |a| = |H|. we can do this, because H is a subgroup of a cyclic group, so is itself cyclic. just knowing that for every g in G, g^{k}= e, doesn't tell you k = |G| (for example, in V, the klein 4-group, it's true that g^{2}= e, for every element g, but |V| = 4, not 2).

you *can* however conclude this, *in the case that G is cyclic* (because then, for a generator x, |x| = |G|).

******

to see specifically what i mean: suppose p = 13, and suppose d = 3.

first let's look at all 3rd powers of non-zero elements:

1^{3}= 1

2^{3}= 8

3^{3}= 27 = 1

4^{3}= 64 = 12

5^{3}= (25)(5) = (12)(5) = 60 = 8

6^{3}= (36)(6) = (10)(6) = 8

7^{3}= (49)(7) = (10)(7) = 5

8^{3}= (64)(8) = (12)(8) = 5

9^{3}= (81)(9) = (3)(9) = 1

10^{3}= (100)(10) = (9)(10) = 12

11^{3}= (121)(11) = (4)(11) = 5

12^{3}= (144)(12) = (1)(12) = 12

so here H = {1,5,8,12}.

now certainly we can take a = 12, m = 4. and it *is* true that 4^{(3)(4)}= 1. but 4 only has order 6 in Z_{13}*:

4^{2}= 3

4^{3}= 12

4^{6}= 12^{2}= 1, so that 12 = 4^{3}only has order 2 in H. and thus we can only conclude from 12 = 4^{3}that 2 divides (p-1)/d = 4.

on the other hand, choosing 5 or 8 leads to the desired conclusion at once.