I'm not the one who is not sure
Each TIP3055 will control 40 LED's? Each TIP3055 will be powered from 12vdc? Generally, a "standard" LED needs 20ma. The color of the LED determines the Vf (forward voltage) which is in the neighborhood of 2 volts. 4 Red LED's in series would need 8v. The TIP will add≈ 0.5v. So 8 + 0.5 = 8.5v. 12v - 8.5v = 3.5v to drop across the collector resistor. Rc = E/I = 3.5v/20ma = 3.5/0.02 = 175Ω. Use the nearest standard value, 180Ω.
For 40 LED's, you will have 10 groups of 4 LED's and a resistor in parallel, all between +12v and the collector. The current for each LED/res group is 20ma. 10 groups is 200ma. Ic = hfe * Ib. For a saturated TIP3055, hfe≈ 10. 200ma = 10 * Ib. Ib = 20ma. You will need to make sure your driver can produce this 20ma when its output is high. Assuming a 5v driver, Rb (the base resistor) = E/I = (Vdriver - Vbe)/Ib = (5-.65)/.02 = 217.5Ω. Use the nearest standard value, 220Ω.
For only 200ma of collector current, A TIP3055 is huge overkill. A 2N2222 will work nicely.
Many people do not use the saturated hfe for calculating the base resistor. They use min hfe instead. This lowers the base current and raises the base resistor but no longer guarantees ideal operation.
Instead of a BJT, you could use a MOSFET. It's more efficient and does not require any appreciable current from the driver so long as high speed is not a requirement.
Explain about the helper diode. If it's a DC system, it won't need any help.