ABC is a right triangle,

So the radius of the inscribed circle r=(6+8- 10)/2=2.

So S⊙o=4π.

(2) connect OE, OF,

Because AB and AC are tangents of circle O.

So OE⊥AC OF⊥AB

So a, e, d and f are quadrilaterals inscribed in a circle.

So ∠ A+∠ EO ∠ F = 180.

∠EDF= 1/2∠EOF。

So EDF = 90- 1/2 * 88 = 46.

(3) According to (2)

∠EDF=90 - 1/2*∠A

∠FED=90 - 1/2*∠B

∠DFE=90 - 1/2*∠C

So the triangle DEF must be an acute triangle.