How does the inverse product of space permeability and permitivity equals space velocity^2?

Answer 1

not sure about your wording. I've always heard this refered to as permeability of free space and permitivity of free space.

permeability of free space = µ
permitivity of free space =ε = 1 / (c^2 x µ)
where c = speed of light.

so ε x µ = 1 / (c^2 x µ) x µ = 1/c^2

so 1/(ε x µ) = c^2

is that what you're asking?