On the Heisenberg uncertainty relation Are there fundamental limits on the accuracy for measuring both position $q$ at time $t$ and momentum $p$ at time $t+\Delta t$, with tiny $\Delta t$?
If yes, why?
If no, why can't one then measure (in principle) both $q$ and $p$ arbitrarily well at the same time $t$ (which is not allowed by Heisenberg's uncertainty relation), by taking $\Delta t$ sufficiently small and noting that any measurement takes time?
 A: There should be fundamental limitations, but not in a way you'd expect. The uncertainty principle is essentially a reflection of the non-commutativity of the two quantities $x$ and $p$.
According to the Copenhagen interpretation of QM, once you make an exact measurement of the position $x$ at time $t$ (the first measurement can, in itself, be made arbitrarily exact), the wave function collapses to a particular position state, which here would be a dirac delta at $x = x_0$ (say)
$$ |\Psi(t_+)\rangle = |x_0\rangle $$
Assuming we chose the $\Delta t$ to be arbitrarily small so that the state on which one makes the next observation (of $p$) is the same as the one $|\Psi(t_+)\rangle$. In the momentum eigen-basis, this wave function can be written as
\begin{align}
|\Psi\rangle &= |x_0\rangle,\\
&= \int dp ~|p\rangle \langle p| x_0\rangle, \\
&= \int dp ~|p\rangle e^{-ipx_0/\hbar}.
\end{align}
Since the number $e^{-ipx_0/\hbar}$ is a just a phase factor, the probability for the particle to be in any momentum state (or to possess any momentum value) is uniform over all $p$ values. One can as well say that momentum is not well defined right after measurement of position. I hope this answers your question.
If $\Delta t$ is non-zero yet small, uncertainty still shows up and can be seen through either evolving the state (Schrodinger picture) or the operators (Heisenberg picture). In the latter approach, we have a time dependent operator given as
$$ p(\Delta t) = e^{iH \Delta t/\hbar} p(0) e^{-iH \Delta t/\hbar} $$
For small $\Delta t$, $p(\Delta t)$ can be approximated through a truncated Hausdorff expansion. We can quantify the uncertainty through the commutator
\begin{align}
C &= [x(0),p(\Delta t)],\\
 &= [x(0),e^{iH \Delta t/\hbar} p(0) e^{-iH \Delta t/\hbar}] ,
\end{align}
which can be evaluated approximately or accurately depending on the complexity of the Hamiltonian $H$.
A: No, there are no particular limits, but this doesn't cheat the uncertainty principle in the way it might seem. First, it's very useful to remember that the uncertainty principle is not a statement about our knowledge of a quantum state or the nature of measurement, it's a statement about the nature of quantum states. A wavefunction with a well-defined position simply does not have a well-defined momentum. 
Let me try to analyze your example. First, you measure the position of the particle extremely accurately. Now it's in a quantum state that has a very well-defined position and a very poorly defined momentum (like a delta function). Then, at a time $\Delta t$ later, you measure the momentum very accurately. Now it's in a state that has a very well-defined momentum but very poorly defined position (like a sine wave). Other than more or less standard limits on measuring things and state evolution, nothing prevents us from doing exactly this. Importantly, at no point in the above procedure does the state violate Heisenberg's principle.
I'm not sure exactly what you mean by phrases like "any measurement takes time" (and I'm not sure I agree) but if you view the above process that the wavefunction must evolve smoothly from $t$ to $t+\Delta t$ and in no sense do we know the position and momentum arbitrarily well at the same time, namely because at no point in time are they (nor can they be) well defined for the particle.
Aside: I find it most intuitive to think about these things and the measurement process in the many-worlds picture rather than starting to think about exactly at what point in time the wave function "collapses" or anything like that. Hope this helps!
A: In my experimentalist's view, if the HUP is about both  
$$\Delta x\Delta p\ge\frac{\hbar}{2}$$
$$\Delta E\Delta t\ge\frac{\hbar}{2}$$
then a very small $\Delta t$ would imply a very large $\Delta E$ and given that, for a particle, measuring the energy constrains the magnitude of the momentum, making $\Delta t$ very small gives a large uncertainty to $\Delta p$.
True, the  the energy/time uncertainty cannot be derived from commutators, and in this link I found some arguments that imply one has to define well what $t$ in $\Delta t$ is.

does it refer to the accuracy of measurement, to the duration of measurement or perhaps to the lifetime of a decaying state.

In general, I think that paradoxes appear when one mixes frames of reference and this might be one of the cases where one has to keep careful track of the measurement context.
A: After you measured position $q$ at time $t$, even the probability distribution of the momentum you will measure at time $t+\Delta t$ will be "mostly unrelated" to the probability distribution of the momentum you would have measured at time $t-\Delta t$. And if you would measure position again at time $t+2\Delta t$, it will be "mostly unrelated" to both the actual measurement outcome and the probability distribution of the position measurement at time $t$.
You might still ask whether there is any limitation to how accurate you can measure momentum at time $t+\Delta t$, but if you measured position from time $t-\Delta T$ to time $t$ and momentum from time $t$ till time $t+\Delta t$, then this probably just boils down to how accurate you can make a momentum measurement in a finite time $\Delta t$. For a quantum object with mass $m$, this is probably given via the relation $E=\frac{p^2}{2m}$ by $\Delta E=\frac{p+\Delta p/2}{m}\Delta p$. Assuming $\Delta E\Delta t \geq\frac{\hbar}{2}$, then $\frac{p+\Delta p/2}{m}\Delta p\Delta t \geq\frac{\hbar}{2}$.
A: Measuring arbitrarily well means observing eigenstates of the observables, so the answer depends on whether the state can evolve from a position eigenstate to a momentum eigenstate in time $\Delta t$.
