5V and 3.3V power supply
Depending on your input voltage, but at that low current, it makes no real difference. You would simply be changing which one gets the bigger energy wastage (in watts).

The LM1117 shown in your picture has a 5mA quiescent current, a minimum load current of 1.7mA, and typically up to 800mA regardless of the package. 15v Max input. You are using 9v, through a diode.

This is simple Ohm's Law calculations for Power (P = I * V)

Since they are Linear regulators, Current in is Current Out. For the first schematic, that means 9v - 0.7v (Diode Drop), 8.3v - 5v = 3.3v voltage drop at the 5v regulator. Then 5v - 3.3v = 1.7v voltage drop at the 3.3v regulator.
Then we can calculate heat dissipation. 1.7v * (30mA Load + 5mA Quiescent) = 0.06W for the 3.3v reg. For the 5v reg, we have to add the 3.3v's current load to the calculation. 3.3v * (8mA Load + 35mA 3.3v's Reg + 5mA Quiescent) = 0.159W. Grand total wasted power of 0.219W, with 75% of it on the 5v regulator.

For the second schematic, its almost the same. 3.3v drop for the 5v regulator, but 8.3v - 3.3v = 5v drop for the 3.3v regulator. Since the 5v regulator doesn't carry the 3.3v's load, it's just 3.3v * (8mA Load + 5mA Quiescent) = 0.043W. The 3.3v regulator though, 5v * (30mA Load + 5mA Quiescent) = 0.175W. Added together, it's 0.218W, with 75% on the 3.3v regulator.

In any case, the 0.218W in heat is nothing to be concern about. The regulators won't even get warm, let alone hot enough for a heatsink.