1, D(- 1, 2). Let the coordinates of point E be (x, y). According to BC coordinates, we can find the straight line BC: 2x+5y+3 = 0. According to the vector |DC|=2 | DE |, we can get the equation: 4 (x+1) 2+(y-2) 2 =13. The coordinates of point e can be obtained by two equations. Then the answer can be found.

2. The vector is parallel to the straight line, and the slope of the straight line is the same as the direction of the vector:11-m) =-m/2, so that m=2 or-1.

3. Take BC as the X axis and the height on the side of BC as the Y axis. Let the length of the triangle be 2a. Then C(a, 0), B(-a, 0), A(0, √3a), e (-2a/3,0) d (-a/3,2 √ 3a/3). Then according to the intersection of two straight lines, the coordinates of point P can be obtained. Then vector BP is perpendicular to vector CD.