I have a big problem and i don't know how to solve it i have no idea
So, let $i_2: X_2\rightarrow X$ an inclusion and $j_1: X\rightarrow (X,X_1)$ we have that $i_{2_*}: H_k(X_2)\rightarrow H_k(X)$ is injective and $j_{1_*}\circ i_{2_*}: H_k(X_2)\rightarrow H_k(X,X_1)$ is an isomorphism, so in the proof they say that because $i_{2_*}$ is injective then we have : $$H_k(X)=\ker j_{1_*}\oplus \text{Im}~ i_{2_*}$$
this is the proposition and the proof please help me:
What book is it you are quoting?
You need to consider the excision property (if you are to follow the given proof), which is given e.g. in Theorem 2.17 and Corollary 2.18 of the following links: http://www.math.toronto.edu/mgualt/MAT1300/Week%206-9%20Term%202.pdf
The idea is to use the 5-lemma to show the desired isomorphism.
Alternatively, one may show directly that $H_k(X)=H_k(X_1)\oplus H_k(X_2)$ simply because $X_1, X_2$ are a disjoint union and every singular simplex $\sigma:\Delta^k \to X$ is such that its image falls within $X_1$ or $X_2$ (use connectedness of the simplex). Likewise with $A$ and $A_i=A \cap X_i.$
Actually, it is all done here (if you use the 5-lemma): Homology of disjoint union is direct sum of homologies
You have $X=X_1 \sqcup X_2$ union of two disjoint subsets. In fact, by excision you have (here $S_*$ denotes the complex of singular chains of a space or of a pair) a canonical morphism
$$S_*(X_2)=S_*(X_2,\emptyset) \to S_*(X, X_1)$$ which is an isomorphism. If you write the exact sequence of relative homology (before taking homology, i.e. at the level of singular chains) you have
$$0\to S_*(X_1) \to S_*(X) \to S_*(X, X_1) \cong S_*(X_2) \to 0,$$ by the above. You can exchange both $X_1$ and $X_2$, and that means that the above exact sequence splits, i.e.
$$S_*(X)=S_*(X_1)\oplus S_*(X_2)$$ (there are other ways to do this, but let's leave it).
Likewise with $A$ and $A_i$. The direct sums you're looking for are there.