解：\((1)\)灯泡电阻\(R_{L}= \dfrac {U^{2}}{P_{L}}= \dfrac {(220V)^{2}}{25W}=1936Ω\)， \(R_{0}\)两端的电压\(U_{0}=U-U_{L}=220V-176V=44V\)， 由于\(R_{0}\)与灯泡\(L\)串联 所以\(R_{0}= \dfrac {U_{0}}{U_{L}}R_{L}= \dfrac {44V}{176V}×1936Ω=484Ω\)． 原电热毯的额定功率\(P_{{额}}= \dfrac { U_{ {额} }^{ 2 }}{R_{0}}= \dfrac {(220V)^{2}}{484\Omega }=100W\)． 答：原电热毯的额定功率为\(100W\)．
\((2)\)改造后的电热毯在低温档时，\(R_{0}\)与\(R_{X}\)串联， 总发热功率\(P= \dfrac {3}{5}P_{{额}}= \dfrac {3}{5}×100W=60W\)．
电路的总电阻\(R= \dfrac {U^{2}}{P}= \dfrac {(220V)^{2}}{60W}≈806.7Ω\)， 电阻丝的阻值\(R_{X}=R-R_{0}=806.7Ω-484Ω=322.7Ω\)．
答：需串联一个\(322.7Ω\)的电阻丝．\((3)\)改造后的电热毯在低温档时，\(R_{0}\)消耗的电能 \(W_{0}=P_{0}t=I^{2}R_{0}t=(\; \dfrac {P}{U})^{2}R_{0}\;t=( \dfrac {60W}{220V})^{2}×484Ω×10×60s=21600J\)． 答：\(R_{0}\)消耗的电能\(21600J\)．
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