We have by the diagram that $h\circ \eta = \xi\circ h'$. Also, $\ker h = \{x\in X| h(x) = 0_y\}$. What can I say about $\ker \eta$? It's $\{y\in Y | \eta(y) = 0\}$. I thought of saying that every element of $Y$ is in the form $h(x)$ for $x\in X$, but this isn't necessarily true since $h$ is not mentioned to be onto, right? So how can I relate $\ker h$ with $\eta$? How can I do the same for $\xi$ and $h'$?
Also, which property of $\xi$ and $\eta$ that are useful so we can say that they create those homomorphisms?
I think the exercise first asks me to show something that will help me find an exact sequence, and somehow it will help me to relate those things, but I'm lost.
I think you have the composition backwards, we have:
$\eta \circ h = h' \circ \xi$
So suppose $x \in \text{ker }h$. This means $h(x) = 0_Y$, and since $\eta$ is a homomorphism, we have $\eta(0_Y) = 0_{Y'}$.
Since $\eta \circ h = h' \circ \xi$, we have $(h'\circ \xi)(x) = h'(\xi(x)) = 0_{Y'}$, so $\xi(x) \in \text{ker }h'$.
This shows $\xi|_{\text{ker }h}: \text{ker }h \to \text{ker }h'$.
Hopefully this gets you started on the other three properties.
We can regard the pair $(\xi,\eta)$ as a "$2$-morphism", that is a morphism between two morphisms (composition of $2$-morphisms being the obvious concatenation of commutative squares). The special property this pair has is that it forms a commutative square with $(h,h')$.
$2$-morphisms between exact sequences are used in many "diagram-chasing" style proofs.