Electronic – essential to make a super capacitor

What you have will work, although D12 is pointless. The problem is that when the cap is discharging onto the 5 V line, there will be a drop across D13. Using a Schottky as you show is a good idea, but the drop will still be around 1/4 volt. Another problem is that the voltage will go lower over time with the amount of charge drained from the cap.

You might consider putting the energy backup capacity before the power supply. That might allow the stored energy to be used more efficiently, and the voltage will stay regulated for a while.

The energy in a capacitor is

$$ E = \frac{1}{2}CV^2 $$

So at 2.7 V in two capacitors, you would have

$$ E = 2\:\text{capacitors} \cdot \frac{1}{2} \cdot 360\:\mathrm F \cdot (2.7\:\mathrm V)^2 = 2624.4 \:\text{joules} $$

The amount of energy you are taking per second is

$$ 0.5\:\mathrm A \cdot 5\:\mathrm V = 2.5\:\mathrm{J/s} $$

So, in a perfect capacitor with a perfect power supply, you could run this for

$$ \frac{2624.4\:\mathrm J}{2.5\:\mathrm{J/s}} = 1049.76\:\mathrm s = 17.496 \:\text{minutes} $$

Note, that this is only with a perfect capacitor. Supercaps tend to have a high series resistance that loses energy. In addition, the regulator has an efficiency that will vary according to the input, likely going down in various ranges, and certainly not really going to 0 Volts. This means you will have to have the amount of energy to maintain the minimum voltage, and add to it the amount you actually consume for the time you want.

To run a 5 V power supply with 90% efficiency at 0.5 A for 2 hours, or 7200 seconds requires a specific amount of energy to the input of the power supply: $$ 5 V \cdot 0.5 A \cdot 7200 s \cdot \frac{1}{0.90} = 20000 J$$ Note that the 90% efficiency effectively increases the amount of energy needed.

In addition, power supplies typically won't run down to 0 V. So there must be extra energy to handle the amount that is never discharged. We'll call the minimum voltage Vmin.

$$ E_{required} = E_{@Vmax} - E_{@Vmin} = \frac{CV_{max}^2}{2} - \frac{CV_{min}^2}{2}$$ $$ E_{required} = \frac{C}{2} \cdot (V_{max}^2 - V_{min}^2 )$$

So for a Vmax = 2.7 V and a Vmin = 0.5V $$ 20000 J = \frac{C}{2} \cdot (7.04 V^2)$$ $$ C = \frac {20000 J \cdot 2}{7.04} = 5681.8 F$$

This could be one huge capacitor or many smaller adding up in parallel.

However, note that I haven't considered any losses due to series resistance. That just adds on the total capacitance needed, but by using multiple parallel capacitors, you tend to reduce the effective resistance as you are putting them in parallel.