How to calibrate an ammeter and voltmeter?
Most analogue devices, shown on your picture, have a calibration screw which makes it quite easy. These screws are given on the right-bottom side of devices. Using them you can calibrate the output by manually turning the indicator when reference current/voltage is supplied. Do this for maximal value (in this case 15 Amps or 500 Volts). You should use the reference source, capable to provide the exact current/voltage with desired accuracy.

There are no such possibilities in most digital meters avalilable on the market (*), so current/voltage dividers methods are used.

Note that you need to have precise equipment for this, as currents flowing through a voltmeter are microampers, and voltage drops on ammeters are at most milivolts in typical solutions.

For an ammeter you should measure its internal resistance (or impedance if it is AC ammeter). Then you add in parallel a resistance (impedance) so that for desired current (in the picture 15 A) flowing through both ammeter and the resistance will be exact. 

Controlling the resistance you add in parallel you can lower the current flowing through the ammeter, so you do it when the ammeter shows higher current than it really is. There is no possibility to calibrate it when it shows too few (you can't increase the current flowing through the ammeter).

This is also valid when you want to scale your ammeter (eg. showing 5 Amps when there really is 20 A).

The background of this is that (\$Rm\$ - ammeter resistance, \$Ra\$ - resistance added) in parallel connection the equivalent resistance is

$$ \frac{1}{R} = \frac 1 {Rm} + \frac 1 {Ra} $$

$$R = \frac {Rm Ra}{Rm+Ra}$$

For current \$I\$ the voltage drop caused on the system is

$$U = I R = I \frac {Rm Ra}{Rm+Ra}$$

The current flowing through the ammeter is

$$Im = \frac U {Rm} = \frac {I \frac {Rm Ra}{Rm+Ra}} {Rm}  =  \frac {I {Ra}} {Rm + Ra}$$

So changing the \$Ra\$ you can change the current flowing through the meter. For example, if \$Ra = Rm\$ you have half of the current flowing through the ammeter.

Doing this, keep in mind that both \$Ra\$ and \$Rm\$ are small (miliOhms). This cause influence of such elements as wires, connections etc., and temperature factor can have an impact.

This is analogous for a voltmeter, however here you add a resistance in series, to lower the voltage. 

There is also no possibility to increase the measured voltage by connecting in series (you need negative resistance).

Further reading about dividers



(*) some more expensive devices allow scaling by multiplying the output by some factor you type, eg. 1.0001, so this is equivalent to the mentioned screw.