answer:A four-digit palindrome reads the same forwards and backwards and has the form abba. - The thousands digit (a) must be 1 since the palindrome must be between 1000 and 2000. - The hundreds digit (b) can be any integer from 0 to 9 (there are 10 choices). - The tens and units digits are determined by the thousands and hundreds digits respectively, as it must mirror the pattern abba. Thus, the number of different palindromes that can be formed is solely dependent on the choices of the hundreds digit, which are 10. Therefore, there are boxed{10} palindromes between 1000 and 2000.

answer:**Analysis** This question examines the method of solving a quadratic inequality. According to the problem statement, - frac{1}{2} and frac{1}{3} are the roots of the equation ax^{2}+bx+2=0. By applying Vieta's formulas, we can find the result. **Solution** Given that - frac{1}{2} and frac{1}{3} are the roots of the equation ax^{2}+bx+2=0, we have begin{cases}- frac{1}{2}+ frac{1}{3}=- frac{b}{a} - frac{1}{2} times frac{1}{3}= frac{2}{a}end{cases}, solving these equations yields a=-12 and b=-2, thus, a+b=-14. Therefore, the answer is boxed{-14}.

answer:```markdown 1. **Identify the properties of the parallelogram:** Since the diagonals of the parallelogram (ABCD) are perpendicular, (ABCD) must be a rhombus (or a square, which is a special type of rhombus). 2. **Introduce the center of the incircle:** Let (O) be the center of the incircle of (ABCD). Because (ABCD) is a rhombus, the diagonals intersect at (O) and they bisect each other at right angles. 3. **Coordinates and angles:** We need to establish that (angle AOE = angle CFO). Let (angle AOE = theta). Drop a perpendicular (OX) from (O) to side (AB). Notice that (angle AOX = frac{B}{2}), therefore (angle XOE = theta - frac{B}{2}). 4. **Tangency condition:** If (EF) is tangent to the incircle at point (Y), it must be that (angle EOY = angle XOE), hence (angle EOY = theta - frac{B}{2}). 5. **Angle analysis using tangency:** Consider (angle BEF = 2(theta - frac{B}{2}) = 2theta - B). From the tangent condition, we deduce that: [ angle BFE = 180^circ - 2theta ] Therefore, we get [ angle CFE = 2theta ] Hence, (angle CFO = theta). 6. **Establish similarity of triangles:** We have established that (angle AOE = angle CFO). Since (ABCD) is a rhombus, (angle EAO = angle OCF). Therefore, triangles (AOE) and (CFO) are similar by AA similarity criterion. Using the properties of similar triangles: [ frac{AE}{AO} = frac{CO}{CF} ] Consequently, [ AE times CF = AO^2 ] 7. **Similar steps for AH and CG:** Similarly, considering triangle (AHO) and (CGO), we get: [ AH times CG = AO^2 ] 8. **Relate AH/AE and CF/CG:** From the established proportionalities, we can write: [ frac{AH}{AE} = frac{CF}{CG} ] This implies that triangles (AHE) and (CFG) are similar by SAS similarity criterion. Conclusion: Since (AHE) and (CFG) are similar, corresponding angles are equal. Therefore, it follows that: [ angle AEH = angle CGF implies EH parallel FG. ] Thus, we have proved that (EH) and (FG) are parallel. (boxed{)}

answer:Given vectors overrightarrow{a}=(4,2) and overrightarrow{b}=(2-k,k-1), If |overrightarrow{a}+overrightarrow{b}|=|overrightarrow{a}-overrightarrow{b}|, then (overrightarrow{a}+overrightarrow{b})^2=(overrightarrow{a}-overrightarrow{b})^2, therefore overrightarrow{a}^2+2overrightarrow{a}cdotoverrightarrow{b}+overrightarrow{b}^2=overrightarrow{a}^2-2overrightarrow{a}cdotoverrightarrow{b}+overrightarrow{b}^2, therefore overrightarrow{a}cdotoverrightarrow{b}=0, therefore 4(2-k)+2(k-1)=0, Solving this, we get k=3. Therefore, the answer is: boxed{3}. From |overrightarrow{a}+overrightarrow{b}|=|overrightarrow{a}-overrightarrow{b}|, we deduce that overrightarrow{a}cdotoverrightarrow{b}=0, and then we set up an equation to find the value of k. This problem examines the application of the dot product in plane vectors and is a basic question.