Poles are given by $(z-1)(z-2)^2=0$

i.e. z=1 is the pole of order 1 & z=2 is the pole of order 2.

Residue at the pole z=1 is given as

$R_1 = lim_{z→1} \,\ ⁡(z-1). \frac{sinπz^2+cosπz^2}{(z-1)(z-2)^2}$

$= lim_{z→1}⁡ \ \frac{sinπz^2+cosπz^2}{(z-2)^2} = \frac{(sinπ+cosπ)}{(1-2)^2} = -1$

Also Residue at the pole z=1 is given as

$R_2 = lim_{z→2} \,\ \ \frac{1}{1!} \frac{d}{dz} (z-2)^2 . \frac{sinπz^2+cosπz^2}{(z-1)(z-2)^2}$

= $ lim_{z→2} \,\ \frac{d}{dz} (\frac{sinπz^2+cosπz^2}{z-1}) = 4π- 1$