Its image passe through points A(-2, 5) and B(-5, p),

∴5=k-2，

∴k=- 10，

∴ The analytical formula of inverse proportional function is y =- 10x.

∴p=- 10-5，

The coordinates of point B are (-5,2).

Let the expression of straight line AB be y=mx+n,

Then 5=-2m+n2=-5m+n,

∴m= 1n=7，

∴ The expression of straight line AB is y = x+7;

(2) Let the expression of CD be y=x+c from AB∨CD in□ABCD.

∴C(0,c),D(-c,0)，

CD = AB，

∴CD2=AB2

∴c2+c2=(-5+2)2+(2-5)2，

∴c=-3，

∴ The coordinates of point C and point D are (0, -3) and (3, 0) respectively;

(3) Let the analytic expression of quadratic function be y=ax2+bx-3, and the image of quadratic function passes through points A and D. 。

∴5=4a-2b-30=9a+3b-3，

∴a= 1b=-2，

∴ The analytic formula of quadratic function is y = x2-2x-3.

Make EF⊥y axis and BG⊥y axis, and the vertical feet are f and g respectively.

OC = OD，BG=CG，

∴∠BCG=∠OCD=∠ODC=45。

∴∠BCD=90，

∠∠DCE =∠BDO，

∴∠ECF=∠BDC，

∴tan∠ECF=tan∠BDC=BCCD