Plane Resistance with Vias
It shouldn't actually increase resistance noticeably. The thing to remember is the via is now part of the plane - you now have a tall cylinder of copper rather than a circle.
If you work out the area of copper you may, depending on the size of the via, find that the copper area of the via is actually larger than that of the copper that would have been in its place. Granted the via will be thinner material, but it is still unlikely to make much of a difference.

In order to calculate it exactly you would need to know the dimensions of the via, the thickness of the plating, and the resistivity of the material (copper, tin, etc.) that makes up the via.

Given we are talking non-RF frequencies, you can consider everything as lumped element and analysis becomes far easier. The copper plane is essentially a resistor, as are the vias. The current through the plane will distribute itself through every possible path proportionally to resistance of that path. If you have copper on both top and bottom connected through vias, and then have a supply at one end and a load at the other, the current will flow though the vias as well as the plane.
You could if you know the resistance of the vias and the plane end up drawing out a very large and complicated tree of resistors to represent it, but in practice you will find that the numbers are so small that rounding errors during the calculations will probably be as much as the difference the vias make to the resistance.

Resistance is given by the equation below (where \$\rho\$ is resistivity, \$L\$ is length, \$A\$ is area, and \$R\$ is the resistance). \$rho\$ can be found from any textbook or internet source, and the other parameters would depend on the via dimensions - unfold a cylinder into a cuboid and calculate its resistance. Calculate the resistance of the copper circle that would otherwise take its place. The second minus the first will give you the difference in resistance that the via makes to the plane.


$$R = \frac{\rho L}{A}$$

No idea. Anything you want - it just involves a bit of basic integration, multiplication, etc.