answer:Solution: (1) From e= frac { sqrt {3}}{3}, we get 2a^2=3b^2. Also, since the line l: y=x+2 is tangent to the circle x^2+y^2=b^2, we find b= sqrt {2} and a= sqrt {3}, thus the equation of the ellipse C_1 is: frac {x^{2}}{3}+ frac {y^{2}}{2}=1. boxed{frac {x^{2}}{3}+ frac {y^{2}}{2}=1} (2) Since MP=MF_2, the trajectory of the moving point M is a parabola with the directrix l_1: x=-1 and focus F_2. Therefore, the equation of the trajectory C_2 of point M is y^2=4x. boxed{y^2=4x} (3) Let Q(0,0), and suppose Rleft( frac { y_{ 1 }^{ 2 }}{4},y_{1}right), Sleft( frac { y_{ 2 }^{ 2 }}{4},y_{2}right), thus overrightarrow {QR}=left( frac { y_{ 1 }^{ 2 }}{4},y_{1}right), overrightarrow {RS}=left( frac { y_{ 2 }^{ 2 }- y_{ 1 }^{ 2 }}{4},y_{2}-y_{1}right), from overrightarrow {QR}cdot overrightarrow {RS}=0, we get frac { y_{ 1 }^{ 2 }( y_{ 2 }^{ 2 }- y_{ 1 }^{ 2 })}{16}+y_{1}(y_{2}-y_{1})=0. Since y_1 neq y_2, simplifying gives y_{2}=-y_{1}- frac {16}{y_{1}}. Thus, y_{ 2 }^{ 2 }= y_{ 1 }^{ 2 }+ frac {256}{ y_{ 1 }^{ 2 }}+32 geq 2 sqrt {256}+32=64 (equality holds if and only if y_1=pm4), since |overrightarrow {QS}|= sqrt {left( frac { y_{ 2 }^{ 2 }}{4}right)^{2}+ y_{ 2 }^{ 2 }}= frac {1}{4} sqrt {( y_{ 2 }^{ 2 }+8)^{2}-64}, and since y_2^2 geq 64, when y_2^2=64, i.e., y_2=pm8, we have the minimum value of |overrightarrow {QS}|=8 sqrt {5}, thus the range of values for |overrightarrow {QS}| is boxed{[8 sqrt {5}, +infty)}.

answer:**Analysis** This problem examines the properties of the logarithmic function. It can be solved by discussing the properties of the logarithmic function according to different cases. **Solution** If a > 1, then the function f(x) is monotonically increasing, f(x)_{min}=fleft( frac{1}{2} right)=log_a frac{3}{2} =1, which gives a= frac{3}{2}. If 0 < a < 1, then the function f(x) is monotonically decreasing, f(x)_{min}=f(1)=log_a 2=1, which gives a=2 (discard this solution), Therefore, the answer is boxed{frac{3}{2}}.

answer:1. **Relative speed calculation**: The two runners move toward each other, so their relative speed is ( 4 + 10 = 14 ) ft/s. 2. **Duration to complete 300 feet**: Since they meet at a relative speed of 14 ft/s, the time until each meeting can be calculated as (frac{300}{14} approx 21.43) seconds. 3. **Number of meetings before 300 feet**: Each time the combined distance covered reaches 600 feet, they would have met, but since they are stopping at 300 feet, we consider how many times 14 ft goes into 300. This is (frac{300}{14} approx 21.43). They meet roughly every (frac{600}{14} approx 42.86) seconds in the full lap scenario, but they stop halfway, so they pass each other approximately (leftlfloor frac{300}{600/14} rightrfloor = leftlfloor 7 rightrfloor = 7) times. 4. **Consider starting point**: Exclude the start, so they meet (7 - 1 = 6) times. Thus, the final answer is (6). The final answer is boxed{The correct choice is (boxed{C) 6}).}

answer:**Analysis** This question examines the determination of parameter values in the truth of universal propositions, which is a basic problem. By using the fundamental knowledge of logic and the idea of separating parameters, the problem is continuously transformed towards simplicity. The key step is f(x)=(ax-1)^2. Finally, it is transformed into the problem of finding solutions for a= frac{1}{x} in (0,1), which is easy. **Solution** Since the proposition "forall x in (0,1), f(x) neq 0" is a false proposition, it is equivalent to its negation "There exists x in (0,1) such that f(x)=0" being true. Since f(x)=(ax-1)^2, it is equivalent to ax=1 having a solution in (0,1), which is equivalent to a= frac{1}{x} having a solution in (0,1). Since x in (0,1) Leftrightarrow frac{1}{x} > 1, the range of a is a > 1. Therefore, the answer is boxed{a > 1}.