Hatcher 2.2.26 Show that if $A$ is contractile in $X$ then $H_n(X,A) =H_n(X) \oplus H_{n-1} (A)$ 
Show that if $A$ is contractible in $X$ then $H_n(X,A) \approx \tilde H_n(X) \oplus \tilde H_{n-1}(A)$

I know that $\tilde H_n(X \cup CA) \approx H_n(X \cup CA, CA) \approx H_n(X,A)$.
And $(X \cup CA)/X = SA$, where $SA$ is the suspension of $A$. So
$H_n((X \cup CA)/X) = H_n(SA)$, where $SA$ is the suspension of $A$. But $SA \simeq A$, and homology is homotopic invariant, we have $H_n((X \cup CA)/X) = H_n(A)$.
I have seen this discussion in one of the post in mathstack but having no clue how to use this in the problem. The long exact sequence I can calculate but why $H_n(X \cup CA, CA) \approx H_n(X,A)$. And where suspension is used and how to split the long exact sequence in the direct sum?

Hatcher suggests to use the fact that $(X\cup CA)/X\simeq SA$: in order to do that you can consider what you obtained in the point (a) of the exercise.
From (a) you know that $A$ is contractible in $X$ iff $X$ is a retract of $X\cup CA$. Since $X$ is a retract of $X\cup CA$ you have that the following sequence splits:
$$0\to \tilde H_n(X)\to \tilde H_n(X\cup CA)\to \tilde H_n(X\cup CA,X)\to 0,$$
hence 
$$\tilde H_n(X\cup CA)\approx \tilde H_n(X)\oplus \tilde H_n(X\cup CA,X).\label{a}\tag{1}$$
Now, 
$$\tilde H_n(X\cup CA,X)\approx \tilde H_n((X\cup CA)/X)\approx \tilde H_n(SA)\approx \tilde H_{n-1}(A).$$
In order to obtain the desired result you just need to recognize $\tilde H_n(X\cup CA)$ as $\tilde H_n(X,A)\approx H_n(X,A)$ and ($\ref{a}$) becomes 
$$H_n(X,A)\approx \tilde H_n(X)\oplus \tilde H_{n-1}(A).$$
