answer:The coordinates of the point symmetric to P(-3,1) with respect to the x-axis are P'(-3,-1). The equation of line P'Q is y= dfrac{-1}{-3-a}(x-a), which simplifies to x-(3+a)y-a=0. The distance d from the center of the circle (0,0) to the line is d= dfrac{|-a|}{sqrt{1+(3+a)^{2}}}=1. Therefore, we find a= -dfrac{5}{3}. Hence, the answer is boxed{-dfrac{5}{3}}. The coordinates of the point symmetric to P(-3,1) with respect to the x-axis are P'(-3,-1), and the equation of line P'Q is y= dfrac{-1}{-3-a}(x-a). By utilizing the condition that the line is tangent to the circle, we can derive the equation and thus conclude. This problem tests the understanding of the relationship between a line and a circle and the application of symmetry, which is considered a basic question.

answer:For p to be true: x^2 + ax + 1 > 0 must always hold for x in mathbb{R}, which gives -2 < a < 2. For q to be true: x + |x - 2a| > 1 must always hold for x in mathbb{R}, which gives a > frac{1}{2}. Since "p or q" is true and "p and q" is false, it means one must be false and the other true. For p true and q false, we get: -2 < a leqslant frac{1}{2}. For p false and q true, we get: a geqslant 2. Therefore, the range of a is -2 < a leqslant frac{1}{2} or a geqslant 2. Thus, the final answer is boxed{-2 < a leqslant frac{1}{2} text{ or } a geqslant 2}.

answer:Part (1): 1. **Induction Hypothesis**: We will use induction on (k) to show that (F_m mid F_{km}) for all (k). - **Base Case**: For (k=0), (F_{0} = 0), which is trivially divisible by any (F_m). For (k=1), (F_m mid F_m) is obviously true. 2. **Inductive Step**: Assume (F_m mid F_{km}) holds for some (k geq 1). We need to prove it for (k+1). - From the recurrence relation of Fibonacci numbers, we know: [ F_{(k+1)m} = F_{km + m} = F_{km-1 + m} F_{m} + F_{km} F_{m+1} ] 3. **Evaluating the Expression**: Both terms of the above expression include factors of (F_m): - (F_{km}) is divisible by (F_m) due to the induction hypothesis. - Therefore, (F_{km-1} F_m) and (F_{km} F_{m+1}) are both divisible by (F_m). 4. **Conclusion**: Thus, [ F_{(k+1)m} = F_{km-1} F_m + F_{km} F_{m+1} ] is divisible by (F_m), establishing that (F_m mid F_{(k+1)m}). Alternative Method for Linear Recurrence Relations: 1. **Consider a General Recurrence**: Any sequence (u_n) defined by (u_{n+2} = au_{n+1} + bu_n) with integer coefficients (a) and (b) and initial conditions (u_0 = 0), (u_1 = 1). - Let (r_1) and (r_2) be the distinct roots of its characteristic polynomial. - The general form of the solution is (u_n = lambda r_1^n + mu r_2^n). 2. **Expressing Terms**: For (m > 0), we aim to show (u_m mid u_{km}) for any (k geq 1). Define (v_k = u_{km}). - This gives us: [ v_k = lambda (r_1^m)^k + mu (r_2^m)^k ] - Therefore, (v_k) satisfies a second-order linear recurrence with characteristic polynomial (x^2 - (r_1^m + r_2^m)x + (r_1 r_2)^m). 3. **Assuming Integer Coefficients**: - Since (r_1 r_2 = -b), the constant term of the polynomial is an integer. - We need to show that the coefficient of (x), (r_1^m + r_2^m), is also an integer. 4. **Induction on Sequence (w_m)**: - Let (w_m = r_1^m + r_2^m). - Starting values: (w_0 = 2), and (w_1 = a), both of which are integers. - The sequence ((w_m)) follows the same recurrence relation as (u_n). 5. **Conclusion**: (w_m) being integers, (v_k) satisfies a linear recurrence with integer coefficients, and (v_0 = 0), (v_1 = u_m) are all divisible by (u_m). Thus, the proof is complete. Part (2): 1. **Divisibility Relation**: - From part (1), we already know (F_{gcd(m,n)} mid gcd(F_m, F_n)). 2. **Greatest Common Divisor Properties**: - Let (d = gcd(m, n)). - By Bézout's Identity, there exist integers (u) and (v) such that: [ um + vn = d ] 3. **Combining Terms Using Fibonacci Linear Combination**: - We want to write (F_d) as a linear combination of (F_{um}) and (F_{vn}). - Utilizing extended definition for negative indices: [ F_d = F_{um-1} F_{vn} + F_{um} F_{vn+1} ] Since (gcd(F_m, F_n)) divides both (F_{um}) and (F_{vn}), it also divides their linear combination (F_d). 4. **Conclusion**: [ gcd(F_m, F_n) mid F_d ] Therefore, [ gcd(F_m, F_n) = F_{gcd(m, n)} ] Thus, the conclusion is: [ boxed{F_{gcd(m, n)} = gcd(F_m, F_n)} ]

answer:If events a and b are independent, then the probability of both events occurring together, denoted as p(a ∩ b), is the product of their individual probabilities. In other words, p(a ∩ b) = p(a) * p(b). We are given that p(a) = 5/7 and p(a ∩ b) = 0.28571428571428575. We can use these values to find p(b). p(a ∩ b) = p(a) * p(b) 0.28571428571428575 = (5/7) * p(b) To find p(b), we divide both sides of the equation by p(a): p(b) = 0.28571428571428575 / (5/7) To perform the division, we can convert the fraction 5/7 to a decimal or multiply both the numerator and denominator of the fraction on the right side by 7/5 to simplify the equation: p(b) = 0.28571428571428575 / (5/7) p(b) = 0.28571428571428575 * (7/5) Now, we can multiply the decimal by the fraction: p(b) = 0.28571428571428575 * 1.4 p(b) = 0.4 Therefore, the probability of event b, p(b), is boxed{0.4} .