$ 解：(1)R_{ L }=\frac {U_{L }^2}{P_{ L }}=\frac {({ 6 }\ \text {V})^2}{{ 3 }\ \text {W}}={ 12 }Ω$
$ (2)由图可知，电源电压U=6\ \text {V}，U=6\ \text {V}时，I_R=0.5\ \text {A}$
$ I_{ L }=\frac {P_{L }}{U_{ L}}=\frac {{ 3 }\ \text {W}}{{ 6}\ \text {V}}={0.5 }\ \text {A}$
$ I=I_R+I_L=0.5\ \text {A}+0.5\ \text {A}=1\ \text {A}$
$ (3)由图知I=0.35\ \text {A}时，U_R=2.5\ \text {V}，$
$ U_{P }=I_{ }R_{ P }={ 0.35 }\ \text {A}×{ 2 }\ Ω={ 0.7}\ \text {V}$
$ U_L=U-U_R-U_P=6\ \text {V}-2.5\ \text {V}-0.7\ \text {V}=2.8\ \text {V}$
$ P_{ L }=U_{ L }I_{ }={ 2.8 }\ \text{V}×{ 0.35 }\ \text{A}={ 0.98 }\ \text{W}$
