for the second part,

The resistivity of a material increases with the increasing temperature and it can be linearly approximated as follows.

rho = rho_0(1+alpha(T-T_0))

or simple we can say Rho is directly proportionate to T

So rho = kT

T in kelvins,

therefore

R = kT * L /A

assuming all the copper wires have the same cross sectioanl area, A then R = C * T * L

Therefore the wire with the greatest ressitance would be one with the greatest T*L value,

The wire with greates T*L value is wire (1) (Which has 305*10)

Therefore the wire with greatest resistance is wire (1)

This question can be approached as below.

The resistance (R) is related to resistivity (rho), length (L) and cross sectional area (A) by the following equation,

R = rho * L /A

first part -- least resistance, since these are same material (Al) we can ignore it and conside about L/A ratio. Therefore the least resistance wire would be the one with the lowest L/A ratio.

We can calculate the L/A ratio as follows for each wire, We can take the diameter*diameter as the area since (A = pi * d^2 /4)

(i) L/A = 5/25 = 0.2

(ii) L/A = 10/9 = 1.11

(iii) L/A = 10/25 = 0.4

(iv)L/A = 5/9 = 0.555

Therefore the least resistance wire will be (i) with 5 cm length and 5 cm diameter.

You are only supposed to ask 1 question at a time. So I will answer only question 1 here and 2nd one I will add my answer.