What is the measure of one interior angle of a regular polygon with 4n+2 number of sides?

This problem is a real sweetheart(not). Very interesting. Glad to have seen it. One way to solve it is to multiply (4n + 2) by some expression y. Then solve the general expression for y.
To find out the normal expression relating the number of sides and the interior angles, you could start with the small central angles of the polygon that add up to 360. That angle = 360/n where n is the number of sides.

Then remember that the 2 base angles of the triangle making up the central angle and the two base angles = 180.

Call one of the base angles = b
2b + 360/n = 180
2b is the measure of the interior angle.

2b = 180 - 360/n
2b = 180(1 - 2/n)
Call 2b = c which is still the interior angle.

c = 180(1 - 2/n)
Now solve for n

cn = 180(n - 2)
cn = 180n - 360
360 = 180n - cn
360 = (180 - c)n
n = 360 / (180 - c) So if you have a square, the interior angle = 90o
n = 360/ (180 - 90)
n = 4 as it should

Suppose the interior angle is 144
Then you get
n = 360 / (180 - 144)
n = 360 / 36
n = 10 as it does.

Now here's the actual proof of what you are asking.

Let n = 4m + 2
Therefore 
4m + 2 = 360 / ( 180 - c) and you are looking to solve for c
(4m + 2) * (180 - c) = 360
180 - c = [360/(4m + 2)] 
c = 180 - [ 360/(4m + 2)]