It's not completely self-evident but it is true that in consistent theories, tachyons have to be scalar particles - much like the Higgs boson when expanded around the maximum of the potential (zero vev) - which obey, of course, Bose-Einstein statistics. (The claim about Fermi-Dirac is just wrong, or was meant to apply to Faddeev-Popov ghosts or similar fields, not physical tachyons.)
In non-interacting string theory, one sees that this conclusion is true because the ground state energy of a single-string Hilbert space is equal to $L_0=-1$ for the scalar tachyon, so any addition of spin - through the string oscillators - raises $L_0$ at least by one, bringing us to the massless or massive level (non-negative $m^2$).
$L_0=-1$ was true for the bosonic string ground state; in the case of the superstring, using the RNS formalism, the ground state has $L_0=-1/2$ but we also have antiperiodic fermions that only raise $L_0$ by $1/2$: that's still enough to show that any addition of spin - which only comes via internal oscillators - brings us to the massless or massive level, above the tachyonic interval.
The scalar character of the tachyon may also be seen in the effective field theory. Dirac or Weyl tachyons are impossible because the Dirac mass term
$$-m \bar \psi \psi$$
has to be Hermitian. It implies that $m$ has to be real, which means that the particle is positively massive. A tachyonic fermion would need an imaginary $m$ but that would produce a non-Hermitian action.
The same is true for spin-one particles. Spin-one particles may only get their mass consistently by the Higgs mechanism: the relevant term arises from the covariant version of the kinetic terms for the Higgs fields:
$$D_\mu \phi^\dagger D^\mu \phi$$
Again, this has to be Hermitean, and if it is Hermitean, $\phi^\dagger \phi$ that is left if the Higgs field has a nonzero vev, is automatically positively definite, which produces the usual mass term
$$m^2 A_\mu A^\mu/2$$
with a positive coefficient.
Concerning a tachyonic vev, well, any vev of a tachyonic field that solves the equations of motion has to be Lorentz-breaking - because it is a non-constant function of spacetime. The spin of a tachyon would just add one more aspect to this story. But it is intuitively natural that the tachyons have to be scalars - the value of the tachyon away from the maximum of the potential measures "how much the instability has already advanced", and this quantity is naturally a scalar.