( For such an example start with any non locally compact space X such that X is Hausdorff. Then the one point compactification Y is weakly Hausdorff but not Hausdorff.)

Thanks for the thoughtful replies. To justify $H \neq K$ in SeqGrp: arxiv.org/abs/1105.6363 constructs When Q=1 a planar continuum X whose fundamental group pi1(X), (with quotient topology inherited from the space of based loops) is the free group on countably many generators, with such topology it is not in TopGrp, but is in SeqGrp. pi1(X) enjoys advertised properties and is finer than $K$.

The free group over countably many generators in the sense of Graev or Markov yields a non-Frechet, sequential, topological group. (Seek the finest topology so that x1,x2,--->1 (or x1, x2,---> the nontrivial element`infinity'))

This answers a different question since a separable sequential group need not be Frechet.

Yes. If X is the fundamental group of the Hawaiian earring (endowed with the quotient topology inherited naturally from the space of based loops). This is arguably overkill but not gratuitously so. For simpler but natural examples, there exist planar continua with countable free fundamental group X so that, topologized with the quotient topology as mentioned, X x X fails to be sequential. Arguably the root pathology lies within the definition of standard product topology, and such phenomena makes the case for the relevance of category theory.