a: the absorbed heat is 2.1× 15 j;
(2) when heating, the resistance R1=U2P heating = (22v) 24w = 121Ω;
p = i2r1 = (ur1+R2) 2r1 during heat preservation;
So 64W=(22V121Ω+R2)2×121Ω
After finishing, R2 = 181.5Ω.
Answer: The resistance values of R1 and R2 are 121Ω and 181.5Ω respectively.