Your statement a) isn't always true.
Water dissociation is represented by:
$$\ce{H2O + H2O <-> H3O+ + OH-}$$
$$ K_\mathrm w=[\ce{H3O+}]\cdot [\ce{OH-}] = 1\times 10^{-14}\ (\textrm{at}\ 25^\circ ~\mathrm C) $$
Note 1: We don't write the $\ce{H_2O}$ activity, since it can usually be rounded to 1 and the ions activities can be rounded to their concentrations.
Note 2: This value was obtained experimentally, considering the concentration of $\ce{H_3O^+}$ and $\ce{OH^-}$ in the medium was the same and measuring the ionization of water.
On a weak acid dissociation:
$$\ce{HA + H2O <-> A- + H3O+}$$
$$K_\mathrm a=\frac{[\ce{A^-}]\cdot [\ce{H3O^+}]}{[\ce{HA}]}$$
On its conjugate base:
$$\ce{A- + H2O <-> HA + OH-}$$
$$K_\mathrm b=\frac{[\ce{HA}]\cdot [\ce{OH^-}]}{[\ce{A^-}]}$$
If we 'add' both reactions we expect that both equilibriums will happen in the mixture, so we have:
$$\ce{HA + H2O <-> A- + H3O+}\ \ \ \ (K_\mathrm a)$$
$$\ce{+}$$
$$\ce{A- + H2O <-> HA + OH-}\ \ \ \ (K_\mathrm b)$$
$$\ce{============================}$$
$$\ce{H2O + H2O <-> H3O+ + OH-}\ \ \ \ (K_\mathrm w)$$
$$K_\mathrm b\cdot K_\mathrm a=[\ce{H3O+}] \cdot [\ce{OH-}] \cdot ([\ce{A-}]\cdot [\ce{HA}])/([\ce{A-}] \cdot [\ce{HA}]) = [\ce{H_3O^+}]\cdot [\ce{OH^⁻}] = K_\mathrm w$$
$$K_\mathrm b=K_\mathrm w/K_\mathrm a~~~~~~ K_\mathrm b=10^{-14}/K_\mathrm a\ \ \textrm{at}\ 25^\circ ~\mathrm C$$
Another way to write this:
\begin{align}-\log(K_\mathrm b)&=-\log(10^{-14}K_\mathrm a) \\ \implies \mathrm pK_\mathrm b &= -\log(10^{⁻14}) - (-\log(K_\mathrm a))\\ \implies \mathrm pK_\mathrm a+\mathrm pK_\mathrm b &=14\;.\end{align}
This is the relation between a conjugate base strength and its acid strength. A very strong acid has a weak conjugate base, but a weak acid doesn't necessarily have a very strong base. Say you have a $\mathrm pK_\mathrm a=5$, which is a weak acid, with $K_\mathrm a=1\times 10^{-5}$. The conjugate base would have a $\mathrm pK_\mathrm b=14-5=9$ or a $K_\mathrm b=1\times 10^{-9}$, which is not a strong base.
However, if we have a strong acid, like $\ce{HCl}$ with a $\mathrm pK_\mathrm a=-6.3$ and $K_\mathrm a=10^{6.3}$. Its conjugate base would have a $\mathrm pK_\mathrm b=14-(-6.3)=20.3$ and $K_\mathrm b=10^{-20.3}$ which is a really weak base.
Hopefully I didn't make it even more confusing for you! It's mostly an reaction equilibrium issue.
