Let's say you have a Hamiltonian of the form
$$
H=\sigma_1\otimes\sigma_2\otimes\sigma_2\otimes\ldots\otimes\sigma_n
$$
There's a straightforward circuit construction that lets you implement its time evolution $e^{-iHt}$. The trick is basically to decompose the state that you're evolving into the components that are in the $\pm 1$ eigenspaces of $H$. Then, you apply the phase $e^{-it}$ to the $+1$ eigenspace, and the phase $e^{-it}$ to the $-1$ eigenspace. The following circuit does that job (and uncomputes the decomposition at the end).
I'm assuming the phase gate element in the middle to be applying the unitary
$$
\left(\begin{array}{cc} e^{it} & 0 \\ 0 & e^{-it} \end{array}\right).
$$

In general, if you want to evolve some Hamiltonian $H=H_1+H_2$ where $H_1$ and $H_2$ are of the previous form, then by far the easiest is to decompose the evolution as
$$
e^{-iHt}\approx \left(e^{-iH_1t/M}e^{-iH_2t/M}\right)^M
$$
for some large $M$ (although there are algorithms with much better scaling behaviour), and each of those small steps $e^{-iH_1t/M}$ can be implemented with the previous circuit.

That said, sometimes there are smarter things that you can do. Your extra example,
$$
H=X\otimes Y\otimes\mathbb{I}+Z\otimes\mathbb{I}\otimes Y
$$
is one such case. I'd start by apply the unitary rotation $U=\frac{Z+Y}{\sqrt{2}}$ to qubits 2 and 3. This is the equivalent to the Hadamard gate, but converts $Y$ into $Z$ instead of $X$. Now stop for a moment and think. If qubits 2 and 3 are in 00, then we're applying $(X+Z)$ to qubit 1. For 01, it's $(X-Z)$, for 10 it's $(Z-X)$, and for 11 it's $-(X+Z)$. Next, let's apply controlled-not from qubit 2 to qubit 3. This just permutes the basis elements slightly. It now says that we have to apply the Hamiltonian
$$
(-1)^{x_2}(X+(-1)^{x_3}Z)
$$
to the state of qubit 1, if qubits 2 and 3 are in the states $x_2x_3$. Next, remember that $X+Z=\sqrt{2}H$ (Hadamard, not Hamiltonian), and that $X\sqrt{2}HX=X-Z$. So, that gives us an easy way to convert between the two bits of Hamiltonian. We'll just replace those two $X$s with controlled-nots controlled by qubit 3. Similarly, we can use a circuit identity
where this time we'll replace the $X$s with controlled-nots controlled off qubit 2.

Overall, I believe the simulation looks like
It might look complicated, but there's none of the splitting up into little time steps that accumulate errors as you go along. It won't apply very often, but it's worth being aware of these sorts of possibilities.

What does the square root factor in with a dot mean - a gate ? – Enrique Segura – 2020-02-22T20:40:25.160

@EnriqueSegura exactly the same as the other one you just asked about: a phase gate with the labeled angle of rotation. – DaftWullie – 2020-02-24T12:17:49.913