Series limiting resistor?
For a zener of voltage VZ fed from a Voltage source of Vin via a resistor of Rz with no other load.

Iz = V/R = (Vin-Vz)/ Rz  ... 1
Call this value of Iz Iz_max.

Zener power dissipation = Vz x Iz  ... 2
Usually zeners are rated to dissipate this power in free air - except maybe for unusual package types. 

In this case you do not specify Vin so Rz can only be calculated in terms of the above parameters  

Rearranging (1)
Rz = (Vin-Vz)/Iz  

Vz will vary somewhat with Iz but is "close enough" to rated value for most purposes.

As you draw current for a load Iz will reduce
Iz_variable  =  Iz_max - Iload.
Regulation will stop when Iload >= Iz_max



Using a 9V transformer, bridge rectifier, 1000 uF capacitor.  

Vdc unloaded = VAC_RMS x 1.414 ~= 12.7V
Actual loaded DC depends on transformer and load current, but probably 11 - 12 VDC.
Assume 11 VDC 

For 3V3 zener with Imax = 75 mA max current.
Rz must drop 11-3.3 = 7.7V
Rz = V/I = (11-3.3)/.075 = 7.7/.075 = 103 Ohms.
Say 100 Ohms.
Power dissipation in resistor = v^2/R = (11-3.3)^2/100 = 592 mW.
Use a 1 watt resistor (at least).  

Power in zener with no load =~ V x I = 3.3 x 0.075 =~ 250 mW.
Power lost in resistor is about 2.4 x power dissipated in zener!
Overall efficiency = Vz/Vin = 3.3/11 = 30%.
This is the result of using a linear regulator.
Using a linear regulator IC will NOT improve efficiency at maximum load but will reduce power dissipation at lighter loads. 

For 4.7 V zener at 53 mA.
Rz = V/I = (11-4.7)/.053 = 119 Ohms.
Use 120 Ohms.
Max efficiency = Vz/Vin = 4.7/11 ~= 43%.
Resistor power dissipation = Vr^2/R = (11-4.7)^2/120 = 0.33 watt.
Use 1/2 watt or higher rated resistor.

Using linear regulators will give better regulation and better efficiencies at lower loads.    This is because a zener regulator always draws its design current at a given voltage - either via the zener to ground or into the load.
A linear regulator uses a small amount of current to operate the regulator and then passes only the required current.