answer:Since f(3)=2, f(4)=f(3)+3, f(5)=f(4)+4, ... f(n-1)=f(n-2)+n-2, f(n)=f(n-1)+n-1, By summing these up, we get f(n)=2+3+...+(n-2)+(n-1) = frac {1}{2}(n-2)(n-1+2)= frac {1}{2}(n+1)(n-2) Thus, the final answer is boxed{f(n)=frac {1}{2}(n+1)(n-2)}. To find the value of f(n) for n>4, we need to give the values of f(3), f(4),..., f(n-1), f(n) and then analyze the relationship between each term. Then, we can use the sum of series to solve the problem. This problem tests your understanding of inductive reasoning and sum of series. Analyzing the relationship between each term and identifying the pattern of change between them is the key to solving this problem.

answer:Since A={-2leqslant xleqslant 5} and B={x|m+1leqslant xleqslant 2m-1}, with Bcap A=B, We know that Bsubset A, When B is an empty set, m+1 > 2m-1, Solving for m, we get m < -2, When B is not an empty set, begin{cases} 2m-1leqslant 5 m+1geqslant -2end{cases}, Solving for m, we get (-3leqslant mleqslant 3). In summary: mleqslant 3, Therefore, the range of values for m is (-infty,3]. Hence the answer is: boxed{(-infty,3]}. From the given information, we know that Bsubset A, from which we can find the range of values for m. This problem tests the method of finding the range of values for real numbers and is a basic question. When solving, carefully read the problem and reasonably apply the properties of intersections.

answer:The area of a rectangle is calculated by multiplying its length by its width. For this garden, the length is 12 meters and the width is 5 meters. So, the area is: 12 m * 5 m = 60 m² Therefore, the area of the garden is boxed{60} square meters.

answer:1. **Lemma Application**: We start by applying the lemma to the given problem. Let ABC be a triangle and Delta M_AM_BM_C be the medial triangle of Delta ABC. The tangent to odot(ABC) at C intersects overline{M_BM_C} at P, and overline{M_AM_B} hits odot(ABC) at X. According to the lemma, overline{XA} and overline{XP} are isogonal conjugates with respect to Delta CXM_C. 2. **Intersection Points**: Let overline{XP} cap overline{BC} = Q and overline{AC} cap overline{XM_C} = R. We have the following ratio: [ frac{RX}{RM_C} = frac{M_BX}{M_AM_B} = frac{PX}{PQ} ] 3. **Trigonometric Ratios**: Using trigonometric identities, we can express the ratio frac{PX}{PQ} as: [ frac{PX}{PQ} = frac{CX}{CQ} cdot frac{sin angle PCX}{sin angle QCP} ] Since angle PCX = angle XAR and angle QCP = angle M_CAR, we have: [ frac{PX}{PQ} = frac{CX}{CQ} cdot frac{sin angle XAR}{sin angle M_CAR} ] 4. **Simplifying Ratios**: Given frac{RX}{RM_C} = frac{XA}{AM_C} cdot frac{sin angle XAR}{sin angle M_CAR}, we equate: [ frac{CX}{CQ} = frac{AX}{AM_C} ] 5. **Angle Relationships**: Knowing measuredangle X_AMC = measuredangle XCQ, we conclude that Delta XAM_C cup R stackrel{+}{sim} Delta XCQ cup P. Thus, overline{XA} and overline{XP} are isogonal conjugates with respect to Delta CXM_C. 6. **Returning to the Problem**: Let the tangent to odot(ABC) at A intersect odot(ABC) at V, and let overline{M_AM_B} cap overline{AV} = U. By the Dual of Desargues Involution Theorem on the complete quadrilateral mathbf{Q} equiv {overline{M_AM_B}, overline{AM_B}, overline{AV}, overline{M_AV}}, there is an involution Phi that swaps {(overline{PA}, overline{PM_A}), (overline{PM_B}, overline{PV}), (overline{PU}, overline{PC})}. 7. **Isogonal Conjugates**: From the lemma, we know that overline{PU} and overline{PC} are isogonal conjugates with respect to Delta APM_A. Therefore, from Phi, (overline{PM_B}, overline{PB}) are also isogonal conjugates with respect to Delta APM_A. 8. **Tangent Line**: This implies measuredangle VPM_A = measuredangle APM_C = measuredangle PQM_A, so overline{PV} is tangent to odot(PM_AQ). (blacksquare)