Property of homology I have this proposition, and I have two questions:
1) Why $H_k=\text{Im} i_*\oplus \ker r_*$ ?
2) Why $j_*: \ker r_*\rightarrow H_k(X,A)$ ?

Edit:
For the second, I try the 1th theorem of isomorphism, and I found that there is an isomorphism between $H_k(X)/\ker j∗$ and $H_k(X,A)$, and I don't know how to appear $\ker r∗$ in the place of $H_k(X)/\ker j∗$.
 A: The trick to do (1) goes like this. If you have $x \in H_k(X)$, you apply $ir$ to get $ir(x) \in im i_*$. The cool thing is than you can then just check that $x - irx$ is in the kernel of $r$, and so $x = irx + (x - irx)$ is your decomposition. To show it's direct, you take $0 = iy + z$. Then applying $r$, we get $0 = riy = y$, and so the sum is direct. The same idea lets you write things in split short exact sequences as sums.
For (2), using your observation from the first isomorphism theorem you put in the edit:
\begin{align*}
H_k(X,A) &\cong H_k(X)/(\ker j_*) \\
&= (\ker r_* \oplus im i_*)/(\ker j_*) \\
&= (\ker r_* \oplus \ker j_*)/(\ker j_*) \\
&= \ker r_* \end{align*} Note where we used the exactness condition $im i_* = \ker j_*$.
A: This is an application of a result of homological algebra which is:
If  $0\to N\stackrel{i}{\to} M\stackrel{j}\to P\to 0$ is an exact sequence and $r:M\to N$  such that $r\circ i=id_{N}$ then $\ker r\simeq P$ and $\text{Im}(i)\simeq N$, and $M=\text{Im}(i)\oplus \ker r \simeq N\oplus P$. 
Proof:
Because $i$ is injective we have $\text{Im}(i)\simeq N$.
For $\ker r \simeq P$ it suffice to show that $j:\ker r\to P$ is an isomorphism: if $x\in \ker r$ such that $j(x)=0$ then $x\text{Im}(i)$ hence there exists $y\in N$ such that $x=i(y)$, but $y=r(i(y))=r(x)=0$. for the surjectivity, let $z\in P$, there exists $x\in M$ such that $y=j(x)$ set $x'=i(r(x))$, we have $r(x-x')=r(x)-r(i(r(x)))=r(x)-r(x)=0$, hence $x-x'\in \ker r$ and $j(x-x')=j(x)-j(i(r(x)))=j(x)=z$.
For $M=\text{Im}(i)\oplus \ker r$: let $y\in \text{Im}(i)\cap \ker r$ then there exists $x\in N$ such that $y=i(x)$ , sa $0=r(y)=r(i(x))=x$ we have $y=i(x)=i(0)=0$.
Let $y\in M$, and set $x=r(y)$, we have $x=i(r(y))\in \text{Im}(i)$ we have 
$r(y-x)=r(y)-r(i(r(y)))=r(y)-r(y)=0$, then $z=y-x\in \ker r$, now we get 
$yx+z$ with $x\in \text{Im}(i)$ and $z\in \ker r$.
