Need help calculating resistance for transistor base
Let's design for worst case, that is a good practice.

\$Ic = 133\text{mA}\$

\$h_{FE} = 30\$ # according to the datasheet minimum 30, typically much better; @Ic=100mA

You can calculate Ib now:

\$I_b = \dfrac{I_c}{h_{FE}} = \dfrac{133\text{mA}}{30} = 4.43\text{mA}\$

\$V_{BE,SAT} = 0.95\$ # datasheet, nearest match is 50mA. Maximum value, practical value is probably much lower (0.65V)

Now let's calculate the base series resistance. This is equal to the voltage across the resistor, divided by the current through it. The current through the resistor is the same as the base current. The voltage across it is the rail voltage (5V) decreased by the base-to-emitter voltage of the transistor V(CE,sat).

\$R_B = \dfrac{U_{R_b}}{I_b} = \dfrac{V_{CC} - V_{BE}}{I_B} = \dfrac{5 - 0.95}{4.43/1000} = 913\Omega\$

With all the worst case engineering up to here, for once let's just round it up to the nearest E12 resistor value of 1kΩ (or 820Ω for worst case engineering, it will work with either).