PLEASE I don't understand this question.

  262°

Step-by-step explanation:

The relationship between the angle at B (82°) and the arcs of the circle, CGF and CDF, is that the angle is half the difference of those arcs. Of course the sum of those arcs is 360°, since together they make a full circle. So. we have ...

  CGF +CDF = 360

  (CGF -CDF)/2 = 82

We need to solve this system of equations to find CGF.

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Multiplying the second equation by 2 and adding the first, we get ...

  2((CGF -CDF)/2) +(CGF +CDF) = 2(82) +(360)

  2CGF = 524 . . . . . simplify

  CGF = 262 . . . . . . divide by 2

The measure of arc CGF is 262°.

Alternate solution

Here's another way to get there. Arc CDF is the supplement to angle B, so is ...

  CDF = 180° -∠CBF

Of course, arc CGF is 360° minus arc CDF, so ...

  CGF = 360° -(180° -∠CBF)

  CGF = 180° +∠CBF . . . . . simplify . . . please note this is a general solution

Then ...

  arc CGF = 180° +82° = 262°.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.