answer:1. **Substitution**: Let us use the substitution (a = sin alpha), (b = sin beta), (c = sin gamma), and (d = sin delta), where (alpha, beta, gamma, delta in left[-frac{pi}{2}, frac{pi}{2}right]). This substitution is valid because the sine function maps (left[-frac{pi}{2}, frac{pi}{2}right]) to ([-1, 1]), and the square roots in the equations ensure that (a, b, c, d in [-1, 1]). 2. **Rewrite the equations**: Using the substitution, the system of equations becomes: [ begin{cases} sin alpha - sqrt{1 - sin^2 beta} + sqrt{1 - sin^2 gamma} = sin delta sin beta - sqrt{1 - sin^2 gamma} + sqrt{1 - sin^2 delta} = sin alpha sin gamma - sqrt{1 - sin^2 delta} + sqrt{1 - sin^2 alpha} = sin beta sin delta - sqrt{1 - sin^2 alpha} + sqrt{1 - sin^2 beta} = sin gamma end{cases} ] Since (sqrt{1 - sin^2 x} = cos x), the system simplifies to: [ begin{cases} sin alpha - cos beta + cos gamma = sin delta sin beta - cos gamma + cos delta = sin alpha sin gamma - cos delta + cos alpha = sin beta sin delta - cos alpha + cos beta = sin gamma end{cases} ] 3. **Summing the equations**: Sum all four equations: [ (sin alpha + sin beta + sin gamma + sin delta) - (cos beta + cos gamma + cos delta + cos alpha) + (cos gamma + cos delta + cos alpha + cos beta) = (sin delta + sin alpha + sin beta + sin gamma) ] Simplifying, we get: [ sin alpha + sin beta + sin gamma + sin delta - cos beta - cos gamma - cos delta - cos alpha + cos gamma + cos delta + cos alpha + cos beta = sin delta + sin alpha + sin beta + sin gamma ] [ sin alpha + sin beta + sin gamma + sin delta = sin delta + sin alpha + sin beta + sin gamma ] This equation is always true, so it does not provide new information. 4. **Analyzing individual equations**: Consider the first equation: [ sin alpha - cos beta + cos gamma = sin delta ] Rearrange it to: [ sin alpha - sin delta = cos beta - cos gamma ] Using the identity (cos x = sqrt{1 - sin^2 x}), we get: [ sin alpha - sin delta = sqrt{1 - sin^2 beta} - sqrt{1 - sin^2 gamma} ] 5. **Possible solutions**: To solve this, we need to consider the possible values of (alpha, beta, gamma, delta). One simple solution is to assume all angles are equal: [ alpha = beta = gamma = delta ] Then: [ sin alpha - cos alpha + cos alpha = sin alpha ] This is true for any (alpha). Therefore, one solution is: [ a = b = c = d ] 6. **Verification**: Substitute (a = b = c = d) back into the original equations: [ a - sqrt{1 - a^2} + sqrt{1 - a^2} = a ] This simplifies to: [ a = a ] which is always true. The final answer is ( boxed{ a = b = c = d } ).

answer:Solution: (1) According to the problem, y= frac {12+1.05x+(0.05x^{2}+0.15x)-(10.75-0.8x)}{x} = frac {0.05x^{2}+2x+1.25}{x} (x＞0); (2) From (1), we have y=0.05x+ frac {1.25}{x}+2, According to the arithmetic mean-geometric mean inequality, we know: y=0.05x+ frac {1.25}{x}+2 geq 2 sqrt {0.05x× frac {1.25}{x}}+2=2.5 (thousand yuan), (Equality holds if and only if 0.05x= frac {1.25}{x}, i.e., x=5), Thus, when the car is used for 5 years, the annual average expenditure on the car is the lowest, at boxed{2.5} thousand yuan.

answer:To solve this problem, we start by understanding the rate at which the machine produces cans of soda and then calculate how many such intervals are there in 8 hours. First, we note that the machine produces 30 cans of soda every 30 minutes. Since there are 60 minutes in an hour, this means there are 2 sets of 30 minutes in an hour (because 60 div 30 = 2). Given that Sue works for 8 hours, we calculate the total number of 30-minute intervals in 8 hours by multiplying the number of intervals per hour by the total hours worked: [8 times 2 = 16] This means there are 16 sets of 30 minutes in 8 hours. To find out how many cans of soda are produced in 8 hours, we multiply the number of cans produced in each 30-minute interval by the total number of such intervals: [30 text{ cans} times 16 = 480 text{ cans}] Therefore, the machine can produce boxed{480} cans of soda in 8 hours.

answer:**Analysis** By using the formula for the probability of independent events, we can set up a system of equations to solve for P(A) and P(B), and then find P(overline{A}). This allows us to calculate P(overline{A}B). **Solution** Given that events A, B, and C are mutually independent, we have: [ begin{cases} P(AB)=P(A)P(B)= frac{1}{6} P(overline{B}C)=P(overline{B})P(C)= frac{1}{8} P(ABoverline{C})=P(A)P(B)P(overline{C})= frac{1}{8} end{cases} ] Solving these equations, we find P(B)= frac{1}{2} and P(A)= frac{1}{3}. Therefore, P(overline{A}B)=P(overline{A})P(B)=left[1-P(A)right]P(B)=left(1- frac{1}{3}right)× frac{1}{2}= frac{1}{3}. Hence, the answers are P(B)=boxed{frac{1}{2}} and P(overline{A}B)=boxed{frac{1}{3}}.