1) this is tricky, i would proceed as follows:

look at the number of 7-sylow subgroups. this number is 1+7k for some non-negative integer k, and divides 52 = 4*13. as none of 8, 15 or 22 divide 52, we may safely conclude the 7-sylow subgroup is normal. let's call this subgroup P.

now, by cauchy's theorem, G has a subgroup H of order 13. since P is normal, PH is a subgroup of G of order 91. furthermore, H is normal in PH (since H is also a 13-sylow subgroup of PH, the number of such subgroups can only be 1+13m, and also a divisor of 7, hence m = 0).

so form the coset space G/PH, and let G act on it by left-multiplication. this gives a homomorphism φ:G→S_{4}, with ker(φ) ⊆ PH. i claim that in fact PH = ker(φ).

the only possibilities are |ker(φ)| = 1,7,13, or 91. but 364, 52 and 28 do not divide 4! = 24.

but this means that PH is normal in G.

now, PH is cyclic (by the CRT), and since PH is normal in G, we have, for any g in G, gHg^{-1}is a subgroup of order 13 in PH. but we saw above H is the *unique* subgroup of order 13 in PH, so gHg^{-1}= H, for all g in G, H is normal.

********

your second question is MUCH easier: let H be the subgroup of index 3. since |H| > 1, |G| ≥ 6. but letting G act on the coset space G/H, we get a homomorphism of G → S_{3}.

it is easy to see that this homomorphism is non-trivial, since G does not act trivially on the coset space. hence its kernel is not all of G. if the kernel is {e}, then we must have |G| = 6, and thus G is isomorphic to S_{3}, which is not simple.

otherwise, the kernel is a non-trivial proper subgroup of G, so G is not simple.