To extend Fillet's information... to hit a winner, ΔKE would be Δ1/2mv^2, or a total of mv^2 (if players hit at similar speeds). I see suggestions [1] that typical shots are in the 60-80 mph range in men's tennis, and some winners get near 100 mph.
So .057*30^2 (for 70ish mph) is 50ish Joules per winner. 41 winners would only be 2 kJ or 0.0006 kWh. Even 100 mph for every winner, that's only 5 kJ = 0.001 kWh. So that doesn't work as anything significant.
I'm not sure what Fillet was taking from cyclingweekly (seems to be basically W/kg and to get to kWh, so you'd need to multiply out by mass and then hours expended to be useful)... but alternatively [2] shows 300-600 Calories per hour is a reasonable amount exerted playing tennis. That'd be 1/3 to 2/3 of a kWh each hour per player. So a three hour match easily could burn 3-4 kWh combined between the two players. Can probably get to around 10 kWh in a long pro match, but still an order of magnitude off.
One thing Fillet didn't consider is the total shots in the match. I found http://www.tennisabstract.com/charting/ and looked through quite a few... most I got was the Djokovic-Murray 2012 US Open, with just under 2000 shots (add the listed total of shot types for the two each players). At the 70 mph earlier estimate, we're still only talking about 0.02 kWh. Still no useful contribution (makes sense, considering how light a ball is). Seems we're usually tired at the end from running, not from swinging our arm.
So, I too can find no way to get near the values they show.
The only guess I can think of is that they actually mean kW (power) rather than kWh (energy). I saw rough estimates that the impact of a shot lasts a few milliseconds [3]. That seems reasonable. A 70 mph winner could thus be about 13 J/0.004 seconds, ~ 3 kW. 133 kW total for 41 winners. Finally a number\unit that sounds similar to their answer... though that would vary hugely by speed and contact time, and be tough to reasonably calculate?
Long story short, I'm as confused by it as Fillet.
