Question

Consider two widely separated conducting spheres, 1 and 2, the second having three times the diameter of the first. The smaller sphere initially has a positive charge q = 5.00

Answer 1

Because the spheres are "widely separated" the field of one will not effect the charge distribution of the other and the charge will distribute on each sphere uniformly. When connected with a wire the charge will flow so that each sphere is at the same potential, because the connection makes them a single conductor and conductors are equipotential surfaces.

You can use the formula for the potential of a spherical conductor of radius "R" , kq/R, for each and set them equal;

kq1/R1 = kq2/R2

Given R2 = 3R1 ,gives;

kq1/R1 = kq2/3R1

q1 = q2/3

In addition you know the total charge must equal 5.00