According to my calculations, you can only get about 70 watts out of this supply. Here's how to calculate it:
Think of your supply as a battery of Vs volts with an internal resistance of Rs ohms.
Take the two measurements you made with R1=2.6 ohms and R2=3.6 ohms. This yielded currents I1=4.92 amps and I2=3.93 amps respectively. Now,
I1=Vs/(Rs+R1), I2=Vs/(Rs+R2)
You have 2 equations and 2 unknowns, Vs and Rs.
Using simple algebra, Vs=19.53 volts and Rs=1.37 ohms.
You can verify that your 3rd measurement will give you approximately the same results. You should also get about 19.5 volts if you measure the output with no load.
With an internal resistance of 1.37 ohms, you will get maximum power transfer (load power) when the load is equal to the internal resistance, i.e., Rload=1.37 ohms.
Using Rload=1.37 ohms and calculating the output current, I=Vs/(Rs+Rload), or I=19.53/(1.37+1.37).
I=7.13amps.
Output power P=I^2*R, or P=(7.13^2)*1.37
P=69.6 watts
This won't be exact, but I don't think you'll get 90 watts from this supply. If your optimum load is not 1.37 ohms, you'll get even less than 70 watts.
Bummer.