A reading for it

The base dissociation constant for ammonia ($\ce{NH3}$) is $\text{Kb} = 1.8 \times 10^{-5}$.

The solution contains a weak base ($\ce{NH3}$) and its conjugate acid ($\ce{NH4+}$ from $\ce{NH4Cl}$).
Given concentrations:
$[\ce{NH3}] = 5.00 \text{ M}$
$[\ce{NH4+}] = 0.100 \text{ M}$

Rounding to two decimal places (consistent with log values often used in pH calculations, or 2 significant figures after the decimal point based on the pKb value):

The pH of the solution is $\boxed{10.95}$.

The volume of the solution is 1.00 L.
Initial moles of $\ce{NH3} = 5.00 \text{ M} \times 1.00 \text{ L} = 5.00 \text{ mol}$
Initial moles of $\ce{NH4+} = 0.100 \text{ M} \times 1.00 \text{ L} = 0.100 \text{ mol}$

Adding $\ce{HCl}$ (a strong acid) will react with the weak base ($\ce{NH3}$).

Moles of $\ce{HCl}$ added = $0.020 \text{ mol}$
Since $\ce{HCl}$ is the limiting reactant, it will consume $0.020 \text{ mol}$ of $\ce{NH3}$ and produce $0.020 \text{ mol}$ of $\ce{NH4+}$.
New moles of $\ce{NH3} = 5.00 \text{ mol} - 0.020 \text{ mol} = 4.98 \text{ mol}$
New moles of $\ce{NH4+} = 0.100 \text{ mol} + 0.020 \text{ mol} = 0.120 \text{ mol}$

Since the volume remains 1.00 L:
New $[\ce{NH3}] = 4.98 \text{ M}$
New $[\ce{NH4+}] = 0.120 \text{ M}$

Using $\text{pKb} = 4.7447$ from part a):

Rounding to two decimal places:

The pH of the solution after adding $\ce{HCl}$ is $\boxed{10.87}$.

Initial moles of $\ce{NH3} = 5.00 \text{ mol}$ (from part a)
Initial moles of $\ce{NH4+} = 0.100 \text{ mol}$ (from part a)

Adding $\ce{NaOH}$ (a strong base) will react with the conjugate acid ($\ce{NH4+}$).

Moles of $\ce{NaOH}$ added = $0.020 \text{ mol}$
Since $\ce{NaOH}$ is the limiting reactant, it will consume $0.020 \text{ mol}$ of $\ce{NH4+}$ and produce $0.020 \text{ mol}$ of $\ce{NH3}$.
New moles of $\ce{NH4+} = 0.100 \text{ mol} - 0.020 \text{ mol} = 0.080 \text{ mol}$
New moles of $\ce{NH3} = 5.00 \text{ mol} + 0.020 \text{ mol} = 5.02 \text{ mol}$

Since the volume remains 1.00 L:
New $[\ce{NH4+}] = 0.080 \text{ M}$
New $[\ce{NH3}] = 5.02 \text{ M}$

Using $\text{pKb} = 4.7447$ from part a):

Rounding to two decimal places:

The pH of the solution after adding $\ce{NaOH}$ is $\boxed{11.05}$.