(1) solution: take the midpoint e of AC and connect it with DE and BE.

So be and de are the center line of isosceles right triangle ABC, vertical line, angular bisector and the center line of right triangle ADC with an included angle of 30 degrees respectively.

So BE= 1/2AC.

Angle CBE= 1/2 Angle ABC=45 degrees.

Angle BEC=90 degrees

Angle CAD=30 degrees

Angle ADC=90 degrees

DE=CE= 1/2AC

So the triangle CDE is an isosceles triangle

Because angle ADC+ angle CAD+ angle ACD= 180 degrees.

So the angle ACD=60 degrees

So the triangle CDE is an equilateral triangle.

So angle CDE= angle CED=60 degrees.

Because BE=DE (authentication)

So angle EBD= angle BDE

Because the angle BEC+ CED+ EBD+ BDE= 180 degrees.

So angle EBD= angle BDE= 15 degrees.

Because angle COD= angle BDE+ angle CED= 15+60=75 degrees.

So the degree of angle COD is 75 degrees.

(2) Solution: Because the angle AOD+ the angle COD= 180 degrees (the flat angle is equal to 180 degrees).

Angle COD=75 degrees (solved)

So AOD angle = 105 degrees.

Because angle DBC= angle CBE- angle EBD

Angle CBE=45 degrees (solved)

Angle EBD= 15 degrees (solved)

So DBC angle =30 degrees.

So AOD/ DBC= 105/30=7/2.