(1)设等差数列的公差为d,等比数列的公比为q,
由a1=b1=2,得a4=2+3d,b4=2q3,S4=8+6d,
由条件a4+b4=27,S4-b4=10,
+2n+2-6n+2
故an=3n-1,bn=2n,n∈N*.
(2)由(1)得,Tn=2an+22an-1+23an-2+…+2na1;①;
2Tn=22an+23an-1+…+2na2+2n+1a1;②;
由②-①得,Tn=-2(3n-1)+3×22+3×23+…+3×2n+2n+2 =10×2n-6n-10;
而-2an+10bn-12=-2(3n-1)+10×2n-12=10×2n-6n-10;
故Tn+12=-2an+10bn(n∈N*).