The number of digits is 125, so x = 5...(3 points)

The scores of girls are 1 19, 125, 120+y, 128, 134.

The average value is126.8 =119+125+120+y+128+1345.

Y = 8...(6 points)

(2) Let boys with scores higher than 125 be a 1, a2,

Notes a 1= 134, a2= 137,

Let's assume that the girls with scores higher than 125 are b 1, b2 and b3 respectively.

Notes b 1= 128, b2= 128, b3= 134,

All methods of taking two copies from students with 12(5 points) or above:

(a 1，a2)，(a 1，b 1)，(a 1，b2)，(a 1，b3)，(a2，b 1)，

(a2, b2), (a2, b3), (b 1, b2), (b 1, b3), (b2, B3) * *10, …(8 points)

It happens to be a man and a woman:

(a 1, b 1), (a 1, b2), (a 1, b3), (a2, b 1), (a2, b2), (a2, B3) *.

∵6 10=35

Therefore, the probability that these two students happen to be a man and a woman is 35...( 12).