answer:1. Calculate the sum of the interior angles of a hexagon: [ (n-2) times 180^circ = (6-2) times 180^circ = 4 times 180^circ = 720^circ ] 2. Divide the total by the number of sides to find the measure of each interior angle in a regular hexagon: [ frac{720^circ}{6} = 120^circ ] 3. Therefore, the degree measure of each interior angle in a regular hexagon is: [ boxed{120^circ} ]

answer:Solution: (1) Since f(x) is an odd function defined on mathbb{R}, we have f(0)=0, therefore 1-(k-1)=0, therefore k=2. (2) Since the function f(x)=a^{x}-a^{-x} (a > 0 and a neq 1), and f(1) < 0, therefore a- frac {1}{a} < 0, and since a > 0, therefore 1 > a > 0. Since y=a^{x} is monotonically decreasing, and y=a^{-x} is monotonically increasing, f(x) is monotonically decreasing on mathbb{R}. The inequality becomes f(x^{2}+tx) < f(x-4). therefore x^{2}+tx > x-4, that is x^{2}+(t-1)x+4 > 0 always holds, therefore triangle =(t-1)^{2}-16 < 0, solving this gives -3 < t < 5. (3) Since f(1)= frac {3}{2}, a- frac {1}{a}= frac {3}{2}, that is 2a^{2}-3a-2=0, therefore a=2, or a=- frac {1}{2} (discard this solution). therefore g(x)=2^{2x}+2^{-2x}-2m(2^{x}-2^{-x})=(2^{x}-2^{-x})^{2}-2m(2^{x}-2^{-x})+2. Let t=f(x)=2^{x}-2^{-x}, from (1) we know k=2, hence f(x)=2^{x}-2^{-x}, which is obviously an increasing function. Since x geqslant 1, therefore t geqslant f(1)= frac {3}{2}, Let h(t)=t^{2}-2mt+2=(t-m)^{2}+2-m^{2} (t geqslant frac {3}{2}) If m geqslant frac {3}{2}, when t=m, h(t)_{min}=2-m^{2}=-2, therefore m=2 If m < frac {3}{2}, when t= frac {3}{2}, h(t)_{min}= frac {17}{4}-3m=-2, solving this gives m= frac {25}{12} > frac {3}{2}, discard this solution. In conclusion, m=2. Therefore, the answers are: (1) k=2, so boxed{k=2}. (2) The range of values of t for which the inequality always holds is -3 < t < 5, so boxed{-3 < t < 5}. (3) The value of m is 2, so boxed{m=2}.

answer:Let alpha = 3 + sqrt{5} and beta = 3 - sqrt{5}. Consider the expression: [ N = alpha^{500} + beta^{500} = (3 + sqrt{5})^{500} + (3 - sqrt{5})^{500}. ] Expanding using the binomial theorem and considering symmetry, all terms involving sqrt{5} will cancel out, leaving us with an integer. Next, analyze beta: [ beta = 3 - sqrt{5} = frac{(3 - sqrt{5})(3 + sqrt{5})}{3 + sqrt{5}} = frac{4}{3 + sqrt{5}} < 1, ] which implies 0 < beta^{500} < 1. Since N is an integer and beta^{500} < 1, we have [ N - 1 < alpha^{500} < N, ] which means n = lfloor alpha^{500} rfloor = N - 1. Then, [ f = x - n = alpha^{500} - (N - 1) = 1 - beta^{500}, ] and [ 1 - f = beta^{500}. ] Thus, [ x(1 - f) = alpha^{500} beta^{500} = (alpha beta)^{500} = [(3 + sqrt{5})(3 - sqrt{5})]^{500} = 4^{500}. ] Finally, we have [ boxed{4^{500}}. ]

answer:Solution: The equation kx+y+2k+1=0 represents a family of lines. It can be rewritten as k(x+2)+y+1=0. From this, we can derive begin{cases} left.begin{matrix}x+2=0 y+1=0end{matrix}right.end{cases}, which gives us begin{cases} left.begin{matrix}x=-2 y=-1end{matrix}right.end{cases}, This means the line always passes through the point (-2, -1). Therefore, the answer is: boxed{(-2, -1)}. By directly using the family of lines equation, we can find the fixed point coordinates. This question tests the application of the family of lines equation and basic knowledge.