That 37,600A fault amps available is correct with a 500 kVA 480/277-240 transformer! Say, the %Z of the transformer is 3.2%, the fault at the secondary will be 500,000/(1.732 x 240)/0.033 = 37,600A (at the 240V side)!

If the transformer in question is an MV-LV with secondary voltage at 480/277V, the fault current available would then be 500,000/(1.732 x 480)/0.032 = 18,800A for a transformer with %Z = 3.2%! Your 11,558A number is good with a 4% Z also.

Second, the fault on the primary side depends on the POCOs system impedance and the voltage at the point-of-common-coupling (POCC)! You cannot compute for the available fault at your connection point (primary side) using the your load impedance alone even if your load are purely motors (at 100% motor load). If it were, your motor load's contribution to the system fault would be 500,000/(1.732 x 480)/0.2 = 3,007A only (at the 480V side and assuming you neglect the impedance of the lines and the per unit impedance of the motors set at 20%)!