(a) Wouldn't you expect this result? A pure inductor takes in no mean power over a period of time, so the power taken in by the inductor-resistor combination ought to be the same as that dissipated in the resistor alone.
Looking at this in terms of current and voltage, let the circuit current be $$I(t)=I_0 \cos (\omega t)$$
Then the pd across the resistor is $$V_R(t)=I_0 R \cos (\omega t)$$
and that across the inductor is $$V_L(t)=-I_0 \omega L \sin (\omega t)$$
So for your first connection of the wattmeter the instantaneous power is
$$I(t) V(t) = I(t) V_R(t)= I_0^2 R \cos^2 (\omega t)$$
And for your second connection of the wattmeter the instantaneous power is
$$I(t) V(t) = I(t) [V_R(t) + V_L(t)] = I_0^2 R \cos^2 (\omega t)– I_0^2 \omega L \cos (\omega t) \sin (\omega t)$$
The second term averages to zero over any whole number of cycles, leaving you with just the power in the resistor.
(b) How does the wattmeter work? Essentially the circuit current, $I(t)$ is carried in a low impedance path through the meter between the top two terminals, while the voltage, $V(t)$ is registered between the left hand terminal and the bottom right terminal (between which the meter impedance is high). The meter calculates $I(t)V(t)$ (instantaneous power) at frequent intervals and presents you with its mean value over a period of time. This is the mean power.