Series resonance
A) How can I show that frequency band ∆f=R/(2πL),given that at
resonance frequency ω₀= 2πf₀=(LC)^-⅟2, the impedance is a minimum: Z =R and ω₀L=1/(ω₀C) so that current will be maximum. At frequencies away from f₀ the impedance is greater than Z(minimum) and current falls.

Parallel resonance
B) Using complex impedance notation (j2-1), the impedance Z of circuit containing a capacitor C in parallel with an inductance L of non- negligible internal resistance r, at a frequency f, given by:
1/z=I/V=(jωC+ 1/(r+jωL)) with jωL=2πL being the complex impedance L, 1/((jωC))= 1/((j2πfC)) the complex impedance of C, while V and I are the amplitudes of voltage and current.
So, how can I show by separating the real and imaginary parts of 1/z that the impedance Z has a maximum value given by: z_max=(r2+ω2+L2)/r=L/rC at a frequency given by
ω= √((1/LC+r2/L2) ) which reduces to ω=ω₀=〖(LC)〗^(-!/2) when r^2=<<(LC).

Please post your attempt.

Try using the fact that for the minimum of a function, its first order derivative is zero and its second order derivative is < 0.