When doing calculations regarding diodes (in half wave rectifiers), is the peak value of the sinusoidal source used, or is the rms value used?
For example, in Vdc = 0.318Vs, is Vs the peak voltage, or the rms voltage.


You are saying DC, this means that the signal will never change. This also means that the \$V_{rms}\$ is the same as \$V_{peak}\$. 

Or am I misunderstanding your awkward equation perhaps?

Perhaps you mean \$V_{DC} = 0.318 × V_s(t)\$, in this case, \$V_s(t)\$ is AC if understand it correctly, in order to turn it into DC you need to use the RMS value.

\$V_{DC} = 0.318 × \frac{12}{\sqrt2}\$ V. 

I'm not sure where you are going to use this \$V_{DC}\$ since it's not asked of you in the question anywhere. 

Some extra info regarding RMS. If you'd use an AC voltage source of \$100×\sin(t)\$ V to power a 60 W lamp, then you could instead use a DC voltage source of \$\frac{100}{\sqrt2} ≃ 70.7 \$ V. The lamp will consume 60 W if it's being fed a 100 V sine wave, or a 70.7 V DC voltage. 




  Likewise, how is the PIV determined, e.g. in part (e) of this question:


Regarding PIV, which stands for Peak Inverse Voltage. That means what is the maximum voltage across the diode in the reverse direction? 

Well that would be 12 volt, assuming \$V_s(t) = 12\sin(t)\$. 

It would occur when \$\sin(t) = -1\$ => \$V_s(t) = -12\$ V