1. 2700
Assume that abcd year ef month gh day, abcdefgh are different
2011
Assume e=1, f can only be 0, 1, 2 (only 10, 11, 12 months, no 13 months )
Since e=1, a=2, f can only be 0
g can only be 3, h can only be 0 or 1 (only 30 and 31 days)
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And e=1, f=0, so it is not true. e cannot be 1, then e=0
e=0, a=2, g is 0-3, then g can only be 3 or 0
Assume g=3, Then h is 0 or 1.
Since e=0, h=1
When g=1, h=1, it is the 31st, e=0, month <10, you can only go to the big month 01, 03,05,07,08
1 and 3 have been taken by h and g, so they can only be 05, 07, and August
f can only be 5, 7, and 8 ; f has three possibilities
0, 1, 2, and 3 have been taken by e, h, a, and g, and bcd can only take 4-9.
Take f, 3 first There are three possibilities, one of 578 is missing
Take b, 4-9, one of the 6 numbers is missing 578, ***5 possibilities
Take c, the same b, 5 possibilities minus a certain number that b has been taken, ***4 possibilities
Take d, the same as c, 4 possibilities minus a certain number that c has taken, ***3 possibilities Possible
So when g=3, ***3*5*4*3=180 possibilities
When g=1, 012 has been taken by e, g, a .
bcdfh can take any number from 3 to 9 (choose 5 out of 7 numbers)
b: 7 possibilities
c: 6
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d：5
f：4
h：3
So when g=1, ***7*6*5* 4*3 = 2520 possibilities
To sum up, there are 182520=2700
2. 3985
Suppose this fraction is a/ b (a and b are both natural numbers), then
a/b>1/1993, b<1993a
a/b<1/1992, b>1992a
When a=1, 1992 When a=2, 1992*2 3984 When a>=3, b>1992*3>5976 So the minimum value of b is 3985