(1) Let y represent the time required for the waiter to prepare the tools, and P represent the probability. The distribution list of y is as follows;

Y2346p153102510a indicates the event "the waiter starts to prepare tea-making tools for the third customer within 6 minutes", so event a corresponds to two situations:

① It takes 2 minutes for the first customer to prepare tea-making tools, and 3 minutes for the second customer;

② The time required for the first customer is 3 minutes, and the time required for preparing the first customer is 2 minutes;

∴P(A)=P(Y=2)？ P(Y=3)+P(Y=3)？ P(Y=2)

= 15×3 10+3 10× 15=325;

(2) The value of x is 0, 1, 2,

When X=0, it takes more than 4 minutes to prepare the first customer.

∴p(x=0)=p(y>4)= 1 10；

X= 1 corresponds to 2 minutes for the first customer and more than 2 minutes for the second customer.

Or 3 minutes for the first customer or 4 minutes for the first customer,

∴P(X= 1)=P(Y=2)？ P(Y>2)+P(Y=3)+P( Y=4)

= 15×45+3 10+25=4350;

X=2 corresponds to 2 minutes for preparing tea-making tools for two customers.

∴p(x=2)=p(y=2)p(y=2)= 15× 15= 125；

The mathematical expectation of ∴X is e (x) = 0×1/kloc-0+1× 4350+2×125 = 4750.