answer:Given the problem: Let ( S ) be a subset of the set ({1,2,3,cdots, 1989}) such that the difference between any two elements in ( S ) is neither 4 nor 7. We want to determine the maximum number of elements that ( S ) can contain. 1. **Identify Non-Conflicting Subsets:** Consider the subset ({1, 4, 6, 7, 9}). Let's verify the differences: [ begin{aligned} &|4-1| = 3, quad |6-1| = 5, quad |7-1| = 6, quad |9-1| = 8, &|6-4| = 2, quad |7-4| = 3, quad |9-4| = 5, &|7-6| = 1, quad |9-6| = 3, &|9-7| = 2. end{aligned} ] None of these differences are 4 or 7. 2. **Extending the Subset:** For the next interval of 11, ({12, 15, 17, 18, 20}) follows the same pattern: [ begin{aligned} 12 + 11 &= 23, quad 15 + 11 &= 26, quad 17 + 11 &= 28, quad 18 + 11 &= 29, quad 20 + 11 &= 31. end{aligned} ] Again, none of the differences between any of these pairs are 4 or 7, and they do not conflict with previously included elements. 3. **General Formulation:** We generalize this pattern. From each interval of length 11 ((1-11, 12-22, dots, 1989)), we can select exactly 5 elements as demonstrated. 4. **Calculate Intervals and Maximum Elements:** Since ( 1989 = 11 times 180 + 9 ), there are exactly 180 full groups of 11 elements plus one group with 9 elements: [ 5 text{ elements per group} times 180 text{ groups} = 900 text{ elements} + text{ elements from the last group}. ] From the last group ({1981, 1982, ldots, 1989}), we can similarly pick at most 5 elements that maintain the required difference conditions: [ {1981, 1984, 1986, 1987, 1989} ] 5. **Final Count:** Thus, the maximum number of elements in ( S ): [ boxed{905} ]

answer:First, let's calculate the distance Mickey runs. Since Mickey runs half as many times around the block as Johnny, and Johnny runs around the block 4 times, we have: - Number of times Mickey runs around the block = frac{1}{2} times 4 = 2 times. Given that one time around the block equals 200 meters, the total distance Mickey runs is: - Distance Mickey runs = 2 times 200 = 400 meters. Next, we calculate the distance Johnny runs. Since Johnny runs around the block 4 times, and each time is 200 meters, we have: - Distance Johnny runs = 4 times 200 = 800 meters. To find the total distance run by both Mickey and Johnny, we add the distances they each ran: - Total distance run = 400 + 800 = 1200 meters. Finally, to find the average distance run by Johnny and Mickey, we divide the total distance by the number of people, which is 2: - Average distance run = frac{1200}{2} = 600 meters. Therefore, the average distance run by Johnny and Mickey is boxed{600} meters.

answer:Since for any ninmathbb{N}^*, it satisfies a_{n+2}-a_{n}leqslant 3^{n}, a_{n+4}-a_{n}geqslant 10times3^{n}, we have 10times3^{n}leqslant (a_{n+4}-a_{n+2})+(a_{n+2}-a_{n})leqslant 3^{n+2}+3^{n}=10times3^{n}. Therefore, a_{n+4}-a_{n}=10times3^{n}. Thus, a_{2017}=(a_{2017}-a_{2013})+(a_{2013}-a_{2009})+ldots+(a_{5}-a_{1})+a_{1} =10times(3^{2013}+3^{2009}+ldots+3)+ frac{3}{8} =10times frac{3times(81^{504}-1)}{81-1}+ frac{3}{8}= frac{3^{2017}}{8}. Therefore, the answer is: boxed{frac{3^{2017}}{8}}. For any ninmathbb{N}^*, it satisfies a_{n+2}-a_{n}leqslant 3^{n}, a_{n+4}-a_{n}geqslant 10times3^{n}, we can derive 10times3^{n}leqslant (a_{n+4}-a_{n+2})+(a_{n+2}-a_{n})leqslant 3^{n+2}+3^{n}=10times3^{n}. a_{n+4}-a_{n}=10times3^{n}. Using a_{2017}=(a_{2017}-a_{2013})+(a_{2013}-a_{2009})+ldots+(a_{5}-a_{1})+a_{1} and the formula for the sum of a geometric series, we can obtain the result. This problem tests the understanding of recursive relationships in sequences, the general formula and sum formula of geometric sequences, and the method of cumulative summation. It examines reasoning and computational skills and is considered difficult.

answer:(1) By the sine law, we have 2cos A(sin Bcos C+sin Ccos B)=sin A, Since Ain(0,pi), we have sin Aneq 0, Thus, 2cos A=1, i.e., cos A= frac {1}{2}, Since Ain(0,pi), we have A= frac {pi}{3}. (2) Since cos B= frac {3}{5} and Bin(0,pi), we have sin B= sqrt {1-cos ^{2}B}= frac {4}{5}, Thus, sin 2B=2sin Bcos B= frac {24}{25} and cos 2B=1-2sin ^{2}B=- frac {7}{25}, Hence, sin (B-C)=sin [B-( frac {2pi}{3}-B)]=sin (2B- frac {2pi}{3})=sin 2Bcos frac {2pi}{3}-cos 2Bsin frac {2pi}{3} =-frac {24}{25}times frac {1}{2}-left(-frac {7}{25}right)times frac {sqrt {3}}{2}=boxed{frac {7sqrt {3}-24}{50}}.