Preface
It is being assumed throughout this post that the concentration c equals the activity a. This is of course not true for concentrations above $\approx 10^{-3} \ldots 10^{-2}~\mathrm{mol\,L^{-1}}$ but it is common practice in schools and universities as long as the exercise is not very specifically about activities.
$$\mathrm{pH} = -\lg a(\ce{H+}) \xrightarrow{a=c} \mathrm{pH} = -\lg [\ce{H+}]$$
The other general assumption is that temperature is $22~^\circ \mathrm{C}$ and therefore $k_\text{W} = 10^{-14}~\mathrm{mol^2\,L^{-2}}$.
Given error contours are not "real measurements vs. calculations" but absolute differences between approximation n and the cubic equation that is not more simplified than using concentrations instead of activities.
General
If you have a calculator that can solve cubic equations, you don't need to simplify. In addition you only would have to remember one equation, no matter which pKa or concentration occurs (see above for real world restrictions).
Solve the following equation for x, with $x=[\ce{H+}]$:
$$x = \underbrace{\frac{k_\mathrm{W}}{x}}_{[\ce{OH-}]} + \underbrace{\frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}}_{[\ce{A-}]}$$
or as a single equation (that is harder to remember):
$$x^3+x^2 k_\mathrm{a}-x([\ce{HA}]_0 k_\mathrm{a}+k_\mathrm{W})-k_\mathrm{a} k_\mathrm{W} = 0$$
Results from this equation will be the reference pH values for the following comparison of pH approximations.
To calculate the pH for any monoprotic(?) base, simply replace [$\ce{H+}$] with [$\ce{OH-}$], $k_\mathrm{a}$ with $k_b$ and of course [$\ce{HA}$] with [$\ce{B}$].
Approximation 1 and 2
For very strong acids the approximation is that they are completely dissociated, which makes the resulting equation very easy:
\begin{align}
x &= [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg [\ce{HA}]_0
\end{align}
When am I able to use this approximation with an error that is smaller than $\Delta \mathrm{pH} \leq 0.1$? It's the blue area in the picture below ($\mathrm{pK_c} = -\lg [\ce{HA}]_0)$.
The simple rule is: $6.7 \gtrsim \mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5$.
Including the hydroxide ions from the autoprotolysis of water makes it a little bit more complicated and yields a quadratic equation in x:
\begin{align}
x &= [\ce{OH-}] + [\ce{HA}]_0 = \frac{k_\mathrm{W}}{x} + [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg \left(0.5 [\ce{HA}]_0 + \sqrt{0.25 [\ce{HA}]_0^2 + k_\mathrm{W}} \right)
\end{align}
This formula holds within: $(\mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5) \vee (\mathrm{pK_c} \gtrsim 7.4)$.
It's not surprising, that the range of usefulness increases over the whole are of diluted concentrations. And that the high pKa values are also covers is only due to the fact that the pH is 7 everywhere upon concentrations that are below $10^{-7}\ldots10^{-8}$.
Approximations 3 & 4
The following approximations are used for weak acids. It's the second easiest to remember, so it's used quite often. The assumption is, that $x \gg k_\mathrm{a}$.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a}} = 0.5 (\mathrm{pK_c} + \mathrm{pk_a})
\end{align}
It's application range is: $(\mathrm{pK_c} < 13.7 - \mathrm{pk_a}) \wedge (\mathrm{pK_c} < \mathrm{pk_a} - 0.6)$.
It is obvious, that the approximation is useful in a small area where the acid is not too diluted and it's pKa value is not too low. Seems to fit well for weak acids.
Again, what happens if $k_\mathrm{W}$ gets involved involved?
\begin{align}
x &= \frac{k_\mathrm{W}}{x} + \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a} + k_\mathrm{W}}
\end{align}
Again, the range expands over very weak acids and strong diluted solutions as expected: $(\mathrm{pK_c} < \mathrm{pk_a} - 0.6) \vee (\mathrm{pK_c} > 14.2 - \mathrm{pk_a})$.
Approximation 5
The last approximation is for strong acids that are not diluted. The amount of protons in solution is equivalent to the amount of dissociated acid but this time with taking $k_\mathrm{a}$ into account. As only the influence of the autoprotolysis of water is neglected, it is "the most accurate" approximation for high to medium concentrations. This can easily be seen as it includes the scope of both approximations, 1 and 2. It is again a quadratic equation in x.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}\\
\mathrm{pH} &= -\lg(-0.5 k_\mathrm{a} + \sqrt{0.25 k_\mathrm{a}^2 + [\ce{HA}]_0 k_\mathrm{a}})
\end{align}
It can be used if: $(\mathrm{pK_c} < 6.7) \wedge (\mathrm{pK_c} < 13.7 - \mathrm{pk_a})$.
Now there are all those approximations, with all their respective acceptable ranges but it's quite hard to compare five diagrams with another. I made two further pictures that show 1) which approximation is the best per (pkc,pks)-pair and 2) which approximation is the easiest to remember per (pkc,pks)-pair.
1) The best -- which function yields the lowest error in comparison with the exact equation?
2) The easiest -- which function yields the lowest error in comparison with the exact equation and if there is more than function with $\Delta \mathrm{pH} < 0.1$, which one is the easiest to remember?
So ... the pka ranges kind of resemble when to use which function but in the end it's a little bit more complicated. If $k_\mathrm{W}$ is neglected, the low-error-area (LEA) is quite small, only approx. 5 is able to yield acceptable values for a big range of concentrations and pka values. Approximations 1 and 3 should be used with care, as their LEAs are quite fancy. For approximations 1 and 3 it's often better to switch to their low-concentration counterparts 2 and 4 to be on the safe side.
