The soln in my terms
1) let there are 4 players A,B,C,D
For player A
First card a spade 13/52 second card also spade =12/51 and so on .... so 13x12x11x..1/(52x51x50...40)
and similarly for rest B,C,D so ans is 4 x(13 x12 x11....1) /(52x51x...40)
2)For player A having exactly one ace = 12 non ace and one ace at any location 12 non-ace=48x47x46x...37/52x51....41 one ace =4/40 =1/10 now ace can take any 13 position so
prob of A having exactly one ace =13 (48 x47x 46 x....37x4) /(52x51x50....41x40)
and same for B , C ,D so
ans= {13 (48 x47x 46 x....37x4) /(52x51x50....41x40)}4
for more details refer https://math.la.asu.edu/~quigg/teach/courses/421/2014fall/notes/stp421notes.pdf