Forward voltage is the voltage drop across the LED when driven by the rated current. You must connect the anode to + and the cathode to - for the LED to be forward biased.
LEDs are current operated devices, not voltage. The forward voltage rating says that when the current flows through the LED, then that voltage will appear across the LED.
You have incorrectly calculated the resistor.
You need to use \$R = \frac {V_{supply}-V_f}{I}\$ to calculate the resistor. That's \$R = \frac {5 - 2.8}{0.060} = \frac {1.6}{0.060} = 36 ohms\$ for your case.
Calculate the resistor at the lower forward voltage because that's where you'll get the higher value resistor and the lowest current. You want to stay below the rated forward current. (0.06 A for your LED.)
Note that this is only approximate. The forward voltage of the LED could be anywhere from 2.8 to 3.4 volts. Using the 37 ohm resistor, that would be 60 milliamperes at 2.8 volts or 43 milliamperes at 3.4 volts.
Reverse voltage has nothing to do with the supply voltage at all.
Reverse voltage tells you that if you hook up the LED backwards then it will be OK unless you connect it to more than 5V backwards. More than 5V (cathode to +, anode to -) and the LED will most likely be damaged.
It is purely coincidence that the reverse voltage rating of 5V matches your power supply voltage of 5V. You could use 12V with an appropriate resistor (143 ohms) and it would work just as well.