The white LEDs have a Vf of 3.5 V so you can put 13 in series. However that only leaves a tiny voltage so I would suggest 8 or 9 in series. That leaves 48 - (3.5 * 8) = 20 V across the resistor (16.5 V for 9 LEDs). You need a resistor of 20/0.75 = 26.6666 Ω, so use 27 Ω (22 Ω for the string with 9 LEDs). They need to be rated at 25 W or so. The power dissipated will be 15 W but it never hurts to have a safety margin.
You need 3 strings to get to 25 LEDs (two 8s and a 9) so the total current is 3 * 0.75 A or 2.25 A.
The same calculation applies for the blue LEDs, so another 2.25 A
The 50 W LEDs can be in series, and the voltage will be 32 - 36 V, so 12 to 16 V will be left. You need a resistor of about 5 Ω, rated at 75 or 100 W. That current is 3 A.
The total current from the supply is 3 + 2.25 + 2.25 = 7.5 A, so the power is 360 W.
You could cut the current down a tiny bit by increasing the resistors, or put the small LEDs into 5 strings not 6, with smaller resistors.
On big LED currents like that, more sophisticated systems that control the current are common. They may be more efficient and the current it better controlled. Resistors are not controlling the current, they are just restricting it to about the right amount. A lot of power is wasted in the resistors.
You might get away with 4 strings of the smaller LEDs if you use a proper current control circuit. if you have 13 LEDs the voltage is 45.5, only leaving 2.5 V. That would be enough if you are controlling the current. If you try with a resistor, you will get far too much current if the LED voltage turns out to be 3.4 V not 3.5 V
That power supply on ebay doesn't say what the input voltage is. It might not match the mains voltage where you are. It is probably not very well made.