$ 解:(1)P=5×40\ \text {W}+60\ \text {W}+100\ \text {W}=360\ \text {W}$
$ I_{ }=\frac {P_{ }}{U_{ }}=\frac {{360 }\ \text {W}}{{ 220}\ \text {V}}={ 1.64 }\ \text {A}$
$ (2)应选用直径为0.46\ \text {mm}的熔丝$
$ (3)W_{ }={ (5×0.04+0.06) }\ \text {kW}×{ 30×4 }\ \text {h}+{ 0.1 }\ \text {kW}×{30×24×\frac 13 }\ \text {h}={ 55.2 }\ \text {kW}·\text {h}$
$ (4)要交电费55.2×0.52元=28.7元$