How practical would a car booster using supercapacitors be?
I started this reply expecting the answer to be "not a chance" but a quick look at specs and prices suggests you could do something which was interesting and possible useful to some extent but that its really impractical and certainly cost effective so far and is unlikely to be cost effective for a few cycles of Moore's law yet.

Assume starting requires 500 A at 12v for 1 second.
 That's far too high in many cases - but lower currents for longer to much longer are common, especially on a very cold morning.
 Adjust assumptions to suit.

Energy in a capacitor
\$ = \frac{1}{2}CV^2 \$
\$ = 0.5 \cdot 1 \cdot 144 \$
  for 1 Farad at 12 Volt = Say 70 Watt seconds per Farad.

Car starting = \$ 12V \cdot 500A \cdot 1 \$ second as above
 = 6000 Watt seconds.

So capacitance required to supply this energy at 12V =
\$ \frac{6000}{70} =~ 100 F\$.

Most supercaps are 2.5V to 3.3V rated for technical reasons.

You can buty modules like this 42V 100F unit that measures 550 x 270 x 110 mm and weighs 13 kg. The note it store 88200 Joule so is 88200/6000 ~= 15 times as large in capacity that the single start solution above. 

To build a 12V cap from 3v3 caps requires 4 in series and from 2V5 = 5 in series.  Placing capacitors in series reduces capacitance in inverse proportion to quantity so we would need 400F with 3v3 capacitors and 500F with 2V5 ones.

Murphy being active it would be wise to use say 1000F x 12v = 5 x 200F at 2V5.

At this stage it get interesting as we find that eg Digikey WILL sell you supercaps in this range.

Cost is very roughly 10 cents per Farad so a 200F ~~= $20 an 4 cost $80. Say  $1000.

A look at the specs shows we are not there yet.
 NO max discharge current specified but internal resistance of around 10 milliohm.That's perhaps 200+ Amps at short circuit. Loaded for maximum power transfer at Rload = Rinternal = 10 milliohm say, that's 100+A.
 That's not really grunty enough for car starting, and we have not yet looked at voltage droop to extract energy etc.

Note that at \$ \frac{V}{2} \$ a cap has exhausted 75% of its energy.
 If a cap is double the energy content needed then draining it to 70% will deliver half the internal energy with the other half stored for next time"

Pretty clearly, a 'battery' that is good for one start is not usually useful. 
Much bigger caps at bigger charge are needed. And even then it will not be possible to approach the energy capacity of a battery.

So - not practical yet - but slowly heading there. Maybe 10 years (about 7 Moore's law cycles)

470 - 3300 Farad x 2.5 V cells.

Leakage:

Leakage of the above is 0.5C mA - so for a 200 F cap that's 100 mA leakage. A farad will supply that for 10 seconds and drop a volt.

So a 200F cap will take \$ 2 \cdot 200F = 400 \$ seconds to drop a volt. Needs work! Some will be much better than this.