This doesn't directly address your question, but it does give you a way of thinking about torsion in the cohomology of Lie groups in general.
(This is all coming from Borel and Serre's Sur certains sous-groupes des groupes de Lie, which can be found in Commentarii mathematici Helvetici Volume 27, 1953)
As you mentioned above, every compact lie group is rationally a product of odd spheres. But how many odd spheres? Turns out, if G is compact and rank k, then it is rationally a product of k spheres (of various dimensions).
There is an analogous result for torsion. That is, one can define the 2-group of G to be any subgroup which is isomorphic to $(mathbb{Z}/2mathbb{Z})^n$ or some n. One defines the 2-rank of a group as the maximal $n$ of any 2-group in G. (On can show that for connected $G$, the 2-rank is bounded by twice the rank, and is thus finite).
Just to point out something that really threw me when I first learned of these - while the rank is an invariant of the algebra (i.e., all Lie groups with the same algebra have the same rank), the 2-rank of a Group is NOT an invariant of the algebra. For example, the 2-rank of SU(2) is 1 (in fact, -Id is the UNIQUE element of SU(2) of order 2), while the 2-rank of SO(3) is 2 (generated by diag(-1,-1,1) and diag(-1,1,-1) ). The 2-rank of O(3) is 3 (generated by diag(-1,1,1), diag(1,-1,1), and diag(1,1,-1) ).
Now, given $Tsubseteq G$, the maximal torus, it's clear that simply by taking the maximal 2-group in T, that the 2-rank of G is AT LEAST the rank of G. When is it strictly bigger? Precisely when the group G contains 2-torsion.
The analogous result for p-groups and p-torsion (p any prime) also holds.
In short, to understand the existence of the 5-torsion in $E_{8}$, one need only understand why there is a subgroup isomorphic to $(mathbb{Z}/5mathbb{Z})^nsubseteq E_8$ for some $ngeq 9$.
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