A resistor obeys Ohm's law, and choosing a resistor value is a matter of the permissible voltage and current, and power (voltage drop x current) according to application. For your switched inputs, a pullup resistor ought not
to burn up, nor prevent a logic LOW level suitable
for the 7489 inputs (0.8 volts max) when the switch is closed. Switch
resistance of 0.01 ohms, power dissipation of 0.1 watt per resistor, and the datasheet limit of current from the 7489 input pin in low state (1.6 mA)
all might be important; in practice, the power dissipation dominates, here.
Any value of input pullup resistor greater than 250 ohms will suit for LOW input normal function.
For input HIGH, switch leakage current, switch holdoff voltage, and logic HIGH preferred value (about 3V) don't limit your resistor choice at all. What MIGHT limit it, is the drive impedance with respect to input capacitance of the input
pin; if there are transmission lines connected, a 220 ohm pullup and 330 ohm
pulldown (to ground) gives good drive (circa 130 ohms to 3V), suitable for
the load of the cable. Low impedance also protects against accidental signal pickup (stray capacitance, static discharge).
For LED drive, you do the same thing: compare the output voltage and
current drive (tested up to 16 mA into no more than 0.45V) of the 7489 output pin
with the current and voltage drop of the resistor and LED in both the HIGH
and the LOW condition, and find resistor values that (1) do turn the LED
sufficiently ON when output is low, and (2) don't turn the LED ON when output is high, and (3) don't burn up resistors or logic.
220 ohms works, will allow ~11 mA per LED.
Optional considerations: if logic is also driven from the outputs as well as LEDs,
it, too, will require a current allowance. A shunt resistor in parallel with the LED will ensure lights-off, and may improve the high-side logic margin (because
of the up-to-2V offset from the LED).