answer:(1) Let overrightarrow{AB} = overrightarrow{c}, overrightarrow{AC} = overrightarrow{b}. Connect AG and extend it to intersect BC at M, where M is the midpoint of BC. Thus, overrightarrow{AM} = frac{1}{2}(overrightarrow{AB} + overrightarrow{AC}) = frac{1}{2}(overrightarrow{b} + overrightarrow{c}), overrightarrow{AG} = frac{1}{3}(overrightarrow{b} + overrightarrow{c}). Given overrightarrow{AP} = lambdaoverrightarrow{AB} = lambdaoverrightarrow{c}, overrightarrow{AQ} = muoverrightarrow{AC} = muoverrightarrow{b}. Therefore, overrightarrow{PQ} = overrightarrow{AQ} - overrightarrow{AP} = muoverrightarrow{b} - lambdaoverrightarrow{c}, overrightarrow{PG} = overrightarrow{AG} + overrightarrow{PA} = frac{1}{3}(overrightarrow{b} + overrightarrow{c}) - lambdaoverrightarrow{c} = (frac{1}{3} - lambda)overrightarrow{c} + frac{1}{3}overrightarrow{b}. Since points P, G, and Q are collinear, there exists a real number t satisfying overrightarrow{PG} = toverrightarrow{PQ}, Thus, (frac{1}{3} - lambda)overrightarrow{c} + frac{1}{3}overrightarrow{b} = tmuoverrightarrow{b} - tlambdaoverrightarrow{c}, That is, frac{1}{3} - lambda = -tlambda, and tmu = frac{1}{3}, Eliminating the parameter t gives: frac{1}{lambda} + frac{1}{mu} = 3, Since triangle APQ and triangle ABC share a common angle, frac{S_1}{S_2} = frac{|overrightarrow{AP}| times |overrightarrow{AQ}|}{|overrightarrow{AB}| times |overrightarrow{AC}|} = lambdamu, Given 0 < lambda leq 1, 0 < mu leq 1, thus frac{1}{lambda} geq 1, frac{1}{mu} geq 1, Since frac{1}{lambda} = 3 - frac{1}{mu} leq 2, Therefore, 1 leq frac{1}{lambda} leq 2, Since frac{1}{lambda} + frac{1}{mu} = 3, Therefore, mu = frac{lambda}{3lambda - 1}, Therefore, frac{S_1}{S_2} = lambdamu = frac{lambda^2}{3lambda - 1} = frac{1}{-(frac{1}{lambda})^2 + 3(frac{1}{lambda})} = frac{1}{-(frac{1}{lambda} - frac{3}{2})^2 + frac{9}{4}}, Since 1 leq frac{1}{lambda} leq 2, Therefore, when frac{1}{lambda} = frac{3}{2}, frac{S_1}{S_2} has the minimum value frac{4}{9}, when frac{1}{lambda} = 1 or 2, frac{S_1}{S_2} has the maximum value frac{1}{2}, Therefore, the range of frac{S_1}{S_2} is boxed{[frac{4}{9}, frac{1}{2}]}.

answer:The sum of the distances from the point (5, 8) to each of the two foci is: - Distance to first focus (-3, 0): [sqrt{(5+3)^2 + (8-0)^2} = sqrt{8^2 + 8^2} = sqrt{128} = 8sqrt{2}] - Distance to second focus (3, 0): [sqrt{(5-3)^2 + (8-0)^2} = sqrt{2^2 + 8^2} = sqrt{68} = 2sqrt{17}] Thus, the total distance is [8sqrt{2} + 2sqrt{17}.] This sum represents 2a, the length of the major axis: [2a = 8sqrt{2} + 2sqrt{17}] [a = 4sqrt{2} + sqrt{17}] The distance between the foci, which is the linear eccentricity 2c, equals 6. Using the ellipse property (c = sqrt{a^2 - b^2}): [c = 3] [3 = sqrt{(4sqrt{2} + sqrt{17})^2 - b^2}] [b^2 = (4sqrt{2} + sqrt{17})^2 - 9] Solving for b^2: [b^2 = 41 + 16sqrt{34} - 9] [b^2 = 32 + 16sqrt{34}] Finally, the equation of the ellipse in standard form is: [frac{x^2}{(4sqrt{2} + sqrt{17})^2} + frac{y^2}{(32 + 16sqrt{34})} = 1] Thus, the ordered pair (a, b) is boxed{(4sqrt{2} + sqrt{17}, sqrt{32 + 16sqrt{34}})}.

answer:First, we need the prime factorizations: - (60 = 2^2 cdot 3 cdot 5) - (429 = 3 cdot 11 cdot 13) The greatest common divisor (GCD) can be found by taking the common prime factor with the lowest exponent. The only common prime factor between 60 and 429 is 3. Hence, the GCD is 3. To find the least common multiple (LCM), we take each prime factor present in at least one of the numbers, raised to the highest power it appears as among the numbers: - LCM = (2^2 cdot 3 cdot 5 cdot 11 cdot 13) - LCM = (4 cdot 3 cdot 5 cdot 11 cdot 13) - LCM = (8580) Sum of the GCD and LCM = (3 + 8580 = 8583). Therefore, the sum of the greatest common divisor and the least common multiple of 60 and 429 is boxed{8583}.

answer:To find the highest value of 'a' such that the number 365a16 is divisible by 8, we need to understand the divisibility rule for 8. A number is divisible by 8 if the last three digits of the number form a number that is divisible by 8. The last three digits of the number 365a16 are 'a16'. We need to find the largest digit 'a' (0-9) such that 'a16' is divisible by 8. Let's check the divisibility by 8 for each possible value of 'a' starting from 9 and going down to 0: - For a = 9, the last three digits are 916, which is not divisible by 8. - For a = 8, the last three digits are 816, which is divisible by 8 (816 ÷ 8 = 102). Since 8 is the largest digit we can use for 'a' to make the number divisible by 8, the highest value of 'a' is boxed{8} .