(1) Regular triangular prism ABC-a 1 b1each side is equal in length, and it is set to1.

Then ab = AC = BC = bb1= cc1= b1c1=1,

D is the center of the side BCC 1B 1, that is, D is the intersection of diagonal BC 1 and CB 1.

Let AE⊥BC be e, then BE=CE= 1/2, AE=( 1/2)∠3, even DE.

∵ plane BCC 1B 1⊥ plane, ∴DE⊥BC plane, and DE=BC/2= 1/2.

∠ADE is the angle between AD and the plane BCC 1B 1,

∫tan∠ADE = AE/DE =[( 1/2)√3]/( 1/2)=√3，

∴∠ADE=60

The angle between AD and BCC 1B 1 plane is 60.

(2)∠DAE is the angle between AD and plane ABC,

∫tan∠DAE = DE/AE =( 1/2)/[( 1/2)√3]= 1/√3，

∴∠DAE=30，

The angle between AD and plane ABC is 30.