answer:Solution: (1) The smallest positive period of the function f(x) = sin(2omega x + varphi) - 1 (omega > 0, |varphi| < frac{pi}{2}) is frac{2pi}{2omega} = frac{pi}{2}, therefore omega = 2, Since the graph of f(x) passes through the point left(0, -frac{1}{2}right), therefore sin varphi - 1 = -frac{1}{2}, therefore varphi = frac{pi}{6}, f(x) = sinleft(4x + frac{pi}{6}right) - 1, Let -frac{pi}{2} + 2kpi leqslant 4x + frac{pi}{6} leqslant frac{pi}{2} + 2kpi (k in mathbb{Z}), we get -frac{pi}{6} + frac{kpi}{2} leqslant x leqslant frac{pi}{12} + frac{kpi}{2} (k in mathbb{Z}), Thus, the intervals of monotonic increase for f(x) are left[-frac{pi}{6} + frac{kpi}{2}, frac{pi}{12} + frac{kpi}{2}right] (k in mathbb{Z}). (2) Translating the graph of function f(x) to the right by frac{pi}{8} units, we get the graph of y = sinleft(4x - frac{pi}{2} + frac{pi}{6}right) = sinleft(4x - frac{pi}{3}right); Stretching the x-coordinates of all points on the graph to twice their original values (the y-coordinates remain unchanged), we get the graph of function y = g(x) = sinleft(2x - frac{pi}{3}right) - 1. In the interval left[0, frac{pi}{2}right], 2x - frac{pi}{3} in left[-frac{pi}{3}, frac{2pi}{3}right], therefore sinleft(2x - frac{pi}{3}right) in left[-frac{sqrt{3}}{2}, 1right], Thus, the range of values for g(x) = sinleft(2x - frac{pi}{3}right) - 1 in the interval left[0, frac{pi}{2}right] is left[-frac{sqrt{3}}{2} - 1, 0right]. If the function F(x) = g(x) + k has exactly two distinct zeros in the interval left[0, frac{pi}{2}right], According to the problem, the graph of g(x) and the line y = -k have exactly two distinct intersection points, And based on sinleft(frac{pi}{3}right) - 1 = sinleft(frac{2pi}{3}right) - 1 = frac{sqrt{3}}{2} - 1, therefore -frac{sqrt{3}}{2} - 1 leqslant -k < 0. Thus, 0 < k leqslant 1 - frac{sqrt{3}}{2}. Therefore, the range of values for k is boxed{0 < k leqslant 1 - frac{sqrt{3}}{2}}.

answer:Let's denote the length of the landscape as ( L ) and the breadth as ( B ). According to the problem, the breadth is 6 times the length, so we can write: [ B = 6L ] The area of the playground is 4200 square meters, which is 1/7th of the total area of the landscape. Therefore, the total area of the landscape is 7 times the area of the playground: [ text{Total area of landscape} = 7 times 4200 ] [ text{Total area of landscape} = 29400 text{ square meters} ] The area of the landscape can also be expressed in terms of its length and breadth: [ text{Area of landscape} = L times B ] Substituting the expression for ( B ) in terms of ( L ) into the area equation, we get: [ 29400 = L times (6L) ] [ 29400 = 6L^2 ] Now, we can solve for ( L ): [ L^2 = frac{29400}{6} ] [ L^2 = 4900 ] [ L = sqrt{4900} ] [ L = 70 text{ meters} ] Now that we have the length, we can find the breadth by multiplying the length by 6: [ B = 6L ] [ B = 6 times 70 ] [ B = 420 text{ meters} ] So, the breadth of the landscape is boxed{420} meters.

answer:(Ⅰ) The derivative of f(x) is f'(x) = (2x + a)e^{3-x} - (x^2 + ax + b)e^{3-x} = -[x^2 + (a-2)x + b - a]e^{3-x}. Setting f'(3) = 0 gives b = -3 - 2a ldots (3). Thus, f'(x) = -(x - 3)(x + a + 1)e^{3-x}. (1) When -a-1 > 3, i.e., a < -4, - For f'(x) > 0, we have 3 < x < -a-1. - For f'(x) < 0, we have x < 3 or x > -a-1. (2) When -a-1 = 3, i.e., a = -4, f'(x) = -(x - 3)^2e^{3-x}, Since -(x - 3)^2 leq 0 and e^{3-x} > 0, It follows that f'(x) = -(x - 3)^2e^{3-x} leq 0 always holds. (3) When -a-1 < 3, i.e., a > -4, - For f'(x) > 0, we have -a-1 < x < 3. - For f'(x) < 0, we have x < -a-1 or x > 3. Summarizing the above: - When a < -4, the intervals of increase for f(x) are (3, -a-1), and the intervals of decrease are (-infty, 3) and (-a-1, +infty). - When a > -4, the intervals of increase for f(x) are (-a-1, 3), and the intervals of decrease are (-infty, -a-1) and (3, +infty). - When a = -4, f(x) is monotonically decreasing on (-infty, +infty). boxed{8} (Ⅱ) From (Ⅰ), when a > 0, f(x) is increasing on the interval (0, 3) and decreasing on the interval (3, 4), Therefore, the range of f(x) on the interval [0, 4] is [min(f(0), f(4)), f(3)], Since f(0) = -(2a + 3)e^3 < 0, f(4) = (2a + 13)e^{-1} > 0, and f(3) = a + 6, The range of f(x) on the interval [0, 4] is boxed{[-(2a + 3)e^3, a + 6]}.

answer:Given that E(X) = 6, we can apply the linearity of the expectation operator: [ begin{aligned} E[3(X - 2)] &= 3E(X - 2) &= 3(E(X) - E(2)) &= 3(E(X) - 2) &= 3(6 - 2) &= 3(4) &= 12. end{aligned} ] Thus, by applying the properties of expectation, we find that E[3(X - 2)] = boxed{12}. Therefore, the correct answer is B.