Magnetic moments of tetrahedral Cobalt (II) and (III) complexes From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$ 

This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling "together"). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases.
The question isn't really clear exactly what complexes you are referring to.  However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin  while the 3.87 value is for high spin (3/2).  I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar? 

For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.    

If those are not the reasons, see Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments Quarterly Review Chemical Society volume 22, pp 457-498.