LowPro proportional valve drive circuit clarification
The figure numbers below refer to the data sheet linked by you.

Yes, with 5V Input the current will be 0.396 A (rounded up). You'll find this value also listed on the valve data sheet table you've provided.

The resistance of the valve is specified at 25 °C and typically 10.1 Ω. This value can vary from valve to valve and will change with rising temperatures. For lack of more exact values we do our calculations with 10.1 Ω. With my rounded value we'll get 4 V over the coil.

Now we can calculate
UCE = Usup - Ucoil - URsense = 8 V - 4 V - 0.396 V = 3.60 V
This is the same value as you have got. With UCE = 3.60 V and IC = 0.396 A we get 1.43 W power dissipation. Some rough calculations showed that the case temperature could reach over 125 °C without a heat sink. Even if still allowed by the specs I would add a small heat sink.

UBE(on) (same as UBE(sat)) = 2.5 V is the value when the transistor is fully conducting and 3 A are flowing through, which is not the case here. With the help of figure 2 you'll find that UBE(sat) at 0.4 A is approximately 1.4 V. Because the transistor isn't fully conductiong this could be even less.

Figure 1 gives us a current gain (hFE) of approximately 1500 at IC = 0.4 A. Now we can calculate:
IB = IC / hFE = 0.396 A / 1500 = 264 μA
The voltage over R1K is then 264 μA * 1 KΩ = 264 mV.
The output voltage of the op-amp will be:
Uopamp = U1K + UBE + Ucoil + URsense = 0.264 V + 1.4 V + 4 V + 0.396 V = 6.06 V

The "missing" 2 Volts to the 8 V supply voltage aren't lost because the op-amp isn't ideal. The output voltage has to be 6.06 V to define the current through the coil. It is a voltage drop necessary for the correct functioning of the circuit. 

As you see these 6.06 V are already higher than 5 V, so you can't use the 5 V as power supply for the op-amp. The output voltage of the op-amp also can't reach the supply voltage (this is the non ideal part ;-). We have to add additional 1.5 V to get the minimum supply voltage of  7.56 V. Round that up and you get the 8 V.

As for the MOSFET thing, I'll make it short here - no.

A functional description of this circuit would be worth an extra answer. In short the 1 Ω resistor is used for current measurement. The voltage over it is used as a negative feedback signal. Because the op-amp can't deliver 0.4 A the transistor is necessary to amplify the current.

One last remark: only a part of the input voltage range 0-5 V is usable to control the valve opening.