You need to be a bit of a sleuth to figure this out. First - you are correct that the main frequency is 40 micro-sec. It's also apparent that the harmonic's frequency is a multiple of that - you can see this because the sum of the two is 0 at 20 micro-sec and 40 micro-sec. If we call the main frequency f1, then the harmonic is f2 = nf1. So let's see what works:

If n = 1, then you would have two sine waves superimposed on each other, resulting in a nice sine wave that is the sum of the two. But that's not what the curve looks like.

If n=2 then at 10 micro-sec you the harmonic wouild be going negative whil te main is ata max, so the slope woukld be negrive. But we see that the sum continues to increase after 10 micro-sec, so n does not equal 2.

If n=3 then at 20 micro-sec both the main and harmonic are crossing from positive to negative, so the slope of the line at t=20 would be negative. But it looks to be essentailly flat at that point. So n does not equal 3.

If n = 4 then there are several things that work out nicely: at t= 10 microsec when the main is at a max the harmonic is going positive, so the sum would continue to increase, which si what we see. At t=20 both main and harmonic are zero, and the slope of the two are going in opposite directions - hence the slope of the sum is near 0, which matches the diagram.

So if we assume that n=4 we can find the amplitude of the main by looking at t= 10, because at that pount the harmonic is 0. We can see from the diagram that the amplitude there is 7.5.

Now to find the amplitude of the harmonic - note that the sum reaches a max of 9 at 12 micro-sec. Hence:

If you solve for A2 you'll find that it's very close to 2.