Which is exactly what they do...
Note "13A cable" has no meaning in this sense. Since we are talking about fault current situations and not continuous current handling capacity. i.e. you will need a flex with at least 1.5mm^2 conductors to carry 13A continuously. However a 0.5mm^2 flex may still be adequately protected by a 13A fuse while carrying 150A of fault current.
What matters, is will the cable survive long enough without damage to open the protective device. To know that you can do some arithmetic to prove it.
For example, say you have a device with 1.8m of lead, and the flex has
0.5mm^2 CSA conductors. Lets assume its protected by a 13A fuse, and the plug is in the furthest socket from the consumer unit on the circuit. To make matters worse, let's assume the circuit itself only just meats the maximum allowable earth loop impedance at that socket for a circuit protected by a B32 MCB.So let's take the circuit earth loop impedance at the socket as 1.37 Ohms[1].
We can take the loop impedance of the flex at around 87 mOhms/m, or 0.16 Ohms total. So if we add those, we get a total of 1.53 Ohms at the appliance end of the "3A" flex.
So if we have a fault right at the appliance end, that will give a prospective fault current of around 230 / 1.53 = 150A
Now we look at the fusing time for a 13A BS1362 fuse[2]. At 150A that's nicely down in the 0.01 second range or less. So the last step is to work out what the minimum cross sectional area of cable is needed to survive that 150A of fault current for the duration. We can get this using the adiabatic equation [3]:
s = sqrt( I^2 x t ) / k
where k is a constant dictated by the flex construction - we will take it as 115 for a PVC flex. Plug the numbers in and we get:
s = sqrt( 150^2 x 0.01 ) / 115 = 0.13 mm^2
Hence we can conclude that our 0.5mm^2 CSA conductor flex is plenty large enough to withstand the fault when protected by a 13A fuse.
[1][2]