There is only one singularity, at z = 2. It is an essential singularity, but there are only even powers of z–2 in the Laurent series. So there is no term in and therefore the residue is 0.
I, myself, would have said that if a point is an essential singularity the function does not HAVE a residue there but I note that Wikipedia asserts that the residue at an essential singularity, is still the coefficient of in the Laurent series only noting that in this case, that is the only way to find the residue, the other formulas, such differentiating n times where n is the order of the pole, not applying. I guess I've been wrong all these years!
Therefore, as Opalg says, the Laurent series for your original function having only even powers, the coefficient of is 0 and the residue is 0.
For your second function, the coefficient of is 1, as you show, and so the residue is 1.
Ahlfors defines it as "the unique complex number R which makes the derivative of a single-valued function in an annulus ." Ahlfors is evasive about whether such a number exists in the case of an essential singularity, commenting that in this case "there is no [...] procedure of any practical value" for computing residues, "and thus it is not surprising that the residue theorem is comparatively seldom used in the presence of essential singularities."