answer:Solution: (1) From |x-m| m^2. Since the solution set of the inequality |x-m| 3, and a-5 leq -1, therefore the range of a is boxed{1 < a leq 4}.

answer:Answer. Any natural number, greater than or equal to 2.Solution. Note that the numbers 21 and 1001 are divisible by 7, and the sums of their digits are 3 and 2, respectively. Therefore, to get a sum of digits equal to an even number n, we need to take a number whose decimal representation consists of frac{n}{2} groups of digits 1001. Accordingly, to get a sum of digits equal to an odd number 2n+1, we need to take a number whose decimal representation consists of frac{n}{2}-1 groups of digits 1001 and one group of digits 21. A sum of digits equal to 1 cannot be obtained, because powers of ten are not divisible by 7.

answer:To solve the given problem, we follow the steps outlined in the solution and provide a detailed breakdown: For part (1) - Finding aGiven that sin A = 3sin B and using the Law of Sines, we can express this relationship in terms of the sides opposite these angles, which gives us:[ frac{a}{sin A} = frac{b}{sin B} ]Given sin A = 3sin B, we substitute to find:[ a = 3b ]Next, we apply the Law of Cosines to find a relationship involving c^2:[ c^2 = a^2 + b^2 - 2abcos C ]Substituting C = frac{pi}{3}, c = sqrt{7}, and a = 3b into the equation, we get:[ 7 = 9b^2 + b^2 - 2 cdot 3b cdot b cdot frac{1}{2} ]Simplifying, we have:[ 7 = 10b^2 - 3b^2 ][ 7 = 7b^2 ][ b^2 = 1 ]Thus, b = 1. Substituting back to find a, we get:[ a = 3b = 3 times 1 = 3 ]So, the value of a is boxed{3}. For part (2) - Finding sin AGiven a = 3, C = frac{pi}{3}, and c = sqrt{7}, we use the Law of Sines to find sin A:[ frac{a}{sin A} = frac{c}{sin C} ][ sin A = frac{a sin C}{c} ]Substituting the given values:[ sin A = frac{3 times frac{sqrt{3}}{2}}{sqrt{7}} ][ sin A = frac{3sqrt{3}}{2sqrt{7}} ][ sin A = frac{3sqrt{21}}{14} ]Therefore, sin A is boxed{frac{3sqrt{21}}{14}}.

answer:-1 .-4+3 a_{1}-2 a_{2}+a_{3} text {. }Substituting x=-1 into the known identity, we getb_{4}=1-a_{1}+a_{2}-a_{3}+a_{4} .Rearranging and simplifying, we havebegin{array}{l}left(x^{4}-1right)+a_{1}left(x^{3}+1right)+a_{2}left(x^{2}-1right)+a_{3}(x+1) =(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+b_{3}(x+1),end{array}which isbegin{array}{l}(x+1)left[(x-1)left(x^{2}+1right)+right. left.a_{1}left(x^{2}-x+1right)+a_{2}(x-1)+a_{3}right] =(x+1)left[(x+1)^{3}+b_{1}(x+1)^{2}+right. left.b_{2}(x+1)+b_{3}right] . text { Hence }(x-1)left(x^{2}+1right)+a_{1}left(x^{2}-x+1right)+a_{2}(x-1)+a_{3} =(x+1)^{3}+b_{1}(x+1)^{2}+b_{2}(x+1)+b_{3} .end{array}Substituting x=-1 again, we getb_{3}=-4+3 a_{1}-2 a_{2}+a_{3} .

answer:# Solution:log _{x}(6 x-5)>2. Domain: x in(5 / 6 ; 1) cup(1 ;+infty) . log _{x} frac{x^{2}}{6 x-5}1end{array} Leftrightarrowleft{begin{array}{c}5 / 60end{array} Leftrightarrowleft{begin{array}{c}5 / 65end{array} Leftrightarrow 5 / 6<x<1right.right.right..2) left{begin{array}{c}1<x<+infty, frac{x^{2}}{6 x-5}<1end{array} Leftrightarrowleft{begin{array}{c}1<x<+infty, x^{2}-6 x+5<0end{array} Leftrightarrowleft{begin{array}{c}1<x<+infty, 1<x<5end{array} Leftrightarrow 1<x<5right.right.right..Answer: x in(2 / 6 ; 1) cup(1 ; 5).

answer:SolutionLet's denote the colors of the main diagonal by the numbers 1, 2, and 3, as shown in the first figure.| 1 | | X || :--- | :--- | :--- || | 2 | || Y | | 3 |Now consider the colors of the two corner squares of the secondary diagonal, denoted by X and Y in the first figure. Since X is in a row that contains the color 1, a diagonal with the color 2, and a column with the color 3, the color X is not in the set {1,2,3}. By the same argument, the color Y is also not in this set. Moreover, since X and Y are on the same diagonal, they must represent different colors. Therefore, we need at least 3+2=5 colors. To ensure that 5 is the minimum number, it suffices to exhibit a configuration with this number of colors. This is done in the next figure, where each number represents a color.| 1 | 3 | 5 || :--- | :--- | :--- || 3 | 2 | 1 || 4 | 1 | 3 |