Answer: D

Video solution:

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Written solution:

The original range is $28-3=25,$ so the new range must be $50.$ To keep the median $8,$ one added number $x$ must be less than $8$ and the other, $y,$ must be greater than $8.$

If $x\ge3,$ the minimum remains $3,$ so the new maximum must be $53.$ The value $x=3$ would change the mode, so $x\le7.$ Hence $x+y\le7+53=60.$

If $x<3$ and $y>28,$ then $y-x=50,$ so $x+y=2x+50\le54.$ If $y\le28,$ the maximum remains $28,$ forcing $x=-22$ and giving an even smaller sum. Therefore no case exceeds $60.$

The values $x=7$ and $y=53$ preserve the mode and median and give range $50,$ so the maximum sum is $60.$

Thus, D is the correct answer.