Hi,
(1) Here the MOSFET's on-resistance is considered to be very small, as mentioned by R_on << R_Load. Consequently, almost the complete volatage, in Fig. 3.3 it is +10 V, is dropping across the load (the lamp). So here the Drain-Source on-resistance is neglected (if the load resistance is significantly larger), and there is no voltage drop across the Drain-Source when the MOSFET is conductive (ON).
If the MOSFET is considered perfectly turned off (non-conductive), you are able to switch between 0 and 10 V and you have your full-swing of the DC voltage, as you can apply the whole +10 V to your load, or none.
(2) I assume here the aim is to point out the advantage of not needing to provide sufficient base current, as it is required for a BJT, and also mentioned in the begining of the section. So for a BJT with an h_fe of ~100, and considering a tenth for dimensioning, you would have to limit/provide a base current of about 10 mA (for the lamp in Fig. 3.3). So as you can see, you even do not need a gate resistor, as the only current flowing towards and from the gate is the charging and discharging current of the gate-capacitance. Note, an actual circuitry would include a small valued resistor to limit the in/out-rush current, to reduce the stress on the MOSFET. So here you have a high impedance input and a low impedance output.
BR