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?

I don't know what notion of equivalence you're using when you write $SA \sim A$, but consider the case where $A = S^1$. Then $SA = S^2$, and this is not homotopy equivalent to $S^1$. 
Perhaps this is where your argument goes off the rails. 
The intuitive argument is this: an element of $H_n(X, A)$ is a chain in $X$ whose boundary maps to $A$. Since $A$ is contractile in $X$, we might as well regard this as a chain in $(X, pt)$, i.e., in the reduced homology of $X$. 
But if $A$ is interesting, then it's possible that two maps into $A$ are fundamentally different...but these are maps of $n-1$-chains into $A$, and that gives you the $H_{n-1}(A)$ factor. 
