= Solution
For sinusoidal steady state with the $e^{i\omega t}$ convention, <complex impedance> is the phasor ratio $Z=\widetilde V/\widetilde I$, encoding both amplitude ratio and phase. From $V=L\,dI/dt$, an ideal inductor has $Z_L=i\omega L$; from $I=C\,dV/dt$, a capacitor has $Z_C=(i\omega C)^{-1}$.
The original diagram has $R_1,L_1$ in series, and $R_4,C_4$ in parallel. Thus
$$
Z_1=R_1+i\omega L_1,\qquad Z_4=\frac{R_4}{1+i\omega R_4C_4}.
$$
With the right supply node as the reference, the two unloaded <voltage divider> outputs are $\widetilde V_A=V_0R_2/(Z_1+R_2)$ and $\widetilde V_B=V_0Z_4/(R_3+Z_4)$. Subtracting gives
$$
\boxed{V_{AB}=V_0\left[\frac{R_2}{Z_1+R_2}-\frac{Z_4}{R_3+Z_4}\right].}
$$
The <series-RL parallel-RC bridge balance> is $Z_1Z_4=R_2R_3$. Equating its real and imaginary coefficients gives
$$
\boxed{R_1=\frac{R_2R_3}{R_4}=25\,\Omega,\qquad
L_1=R_2R_3C_4=0.250\,\mathrm H.}
$$
These ideal components balance the bridge at any frequency, not just the test frequency.
At $\omega=200\,\mathrm{rad\,s^{-1}}$, the current amplitude through $R_2$ is $5/|50+50i|=0.0707\,\mathrm A$. Using peak rather than RMS supply amplitude,
$$
\boxed{\overline P_{R_2}=\frac12|\widetilde I|^2R_2=0.0625\,\mathrm W.}
$$
In the DC steady state the inductor is a short and the capacitor is open. Both dividers have equal resistors, so each midpoint is at2.5V and the bridge stays balanced. The upper-branch current is0.100A, giving $\boxed{P_{R_2,\rm DC}=0.250\,\mathrm W}$.
For the high-input-resistance voltmeter, hold the other components fixed. At the DC balance,
$$
\frac{dV_{AB}}{dR_1}=-\frac{V_0R_2}{(R_1+R_2)^2}=-0.0500\,\mathrm{V\,\Omega^{-1}},\qquad
\boxed{\delta R_1\simeq\frac{0.001}{0.0500}=0.020\,\Omega.}
$$
The <DC bridge resistance resolution> is a local readout resolution, or0.08% of $R_1$, not a total uncertainty allowing for supply, component calibration or voltmeter loading. <University Physics Volume 2 by Samuel J. Ling and others>[§15.2, resistor/inductor/capacitor phasors and peak/RMS distinction]{ref}
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