Calculate homology of $\mathbb{R}P^2$ and $\mathbb{C}P^n$ using map of pairs I have some problems with the following exercises.

Recall that $\mathbb{R}P^2$ can be viewed as $D^2$ with opposite
  points of $S^1 \subset D^2$ identified; the image of $S^1$ is
  $\mathbb{R}P^1$ inside $\mathbb{R}P^2$. Using the map of pairs
  $(D^2,S^1) \to (\mathbb{R}P^2,\mathbb{R}P^1)$ calculate the homology
  of $\mathbb{R}P^2$. Warning: the method is the same for all
  coefficient rings, but the answer will depend on $R$.

Assume that $R = \mathbb{Z}$.
The long exact sequence for topological pair $(D^2,S^1)$ is
$$\cdots\to H_2(S^1)\to H_2(D^2)\to H_2(D^2,S^1)\xrightarrow{\partial_2} H_1(S^1)\to H_1(D^2)\to H_1(D^2,S^1)\xrightarrow{\partial_1} H_0(S^1)\to H_0(D^2)\to H_0(D^2,S^1)\xrightarrow{\partial_0} 0$$
Since homology groups of $D^2$ and $S^1$ are well-known and using exactness of the sequence, we can rewrite it as
$$\cdots\to 0\to 0 \to H_2(D^2,S^1)\xrightarrow{\partial_2} \mathbb{Z}\xrightarrow{\operatorname{id}} \mathbb{Z}\xrightarrow{0} H_1(D^2,S^1)\xrightarrow{\partial_1} \mathbb{Z}\xrightarrow{0} \mathbb{Z}\xrightarrow{\operatorname{id}} H_0(D^2,S^1)\xrightarrow{\partial_0} 0$$
Finally we obtain $H_0(D^2,S^1)= H_1(D^2,S^1) = \mathbb{Z}$, $H_i(D^2,S^1)=0$ for $i \geq 2$.
I can't proceed further because it is unclear to me how should I link this result with homology of $(\mathbb{R}P^2,\mathbb{R}P^1)$ and then homology of $\mathbb{R}P^2$. I guess that since map of pairs $(D^2,S^1)\to(\mathbb{R}P^2,\mathbb{R}P^1)$ is quite good (it is continuous) then it induces isomorphism between homology groups: $H_*(D^2,S^1)=H_*(\mathbb{R}P^2,\mathbb{R}P^1)$. If so, assuming that we already known homology of $\mathbb{R}P^1$, it is possible to write long exact sequence of pair and find homology of $\mathbb{R}P^2$. Am I right?

Using the pair $(\mathbb{C}P^n,\mathbb{C}P^{n-1})$ calculate the
  homology of $\mathbb{C}P^n$ for all $n$.

I don't understand how even to start the solution because the task is to find homology groups $H_n$ for all $n$, so it seems that we known nothing. OK, I can write long exact sequence for $(\mathbb{C}P^n,\mathbb{C}P^{n-1})$, but without any known groups it is impossible to find homology of $\mathbb{C}P^n$. How should it be done?
Any help will be very appreciative.
 A: You didn't use the "map of pairs", just the long exact sequence of homology of the pair $(D^2,S^1)$. Using the map and the naturality of the long exact sequence, we have the diagram below (where we are using $\mathbb{Z}$ coefficients, neglecting orientation signs *- i.e., not specifying generators -, and identifying the homologies while doing so).
$$\require{AMScd}
\begin{CD}
H_2(D^2) @>>> H_2(D^2,S^1) @>>> H_1(S^1) @>>>H_1(D^2) ;\\
 @Vf_*VV @Vf_*VV @Vf_*VV @Vf_*VV\\
H_2(\mathbb{R}P^2) @>>> H_2(\mathbb{R}P^2,\mathbb{R}P^1) @>\partial_*>> H_1(\mathbb{R}P^1) @>>>H_1(\mathbb{R}P^2) ,
\end{CD}$$
where everything to the left (which does not appear above) is zero.
You will now need some result (like computation by degree) to infer that the third $f_*$ has degree $2$. This implies that $\partial_*$ is injective, since the second $f_*$ is an isomorphism (since it is coming from a map of degree $1$ - one way to see this is by using the fact that $(D^2,S^1)$ and $(\mathbb{R}P^2,\mathbb{R}P^1)$ are good pairs, and the map induced by $f$ on the quotient is the identity on the sphere) and therefore $H_2(\mathbb{R}P^2)=0$. The sequence below says that $H_1(\mathbb{R}P^2)=\mathrm{coker}\partial_*$, hence $H_1(\mathbb{R}P^2)=\mathbb{Z}/2\mathbb{Z}$. The remaining computations are trivial.
If you are looking at this with coefficients other than $\mathbb{Z}$, then $\partial_*$ need not be injective, since the image of the third $f_*$ can be $0$. With effect, $f_*$ will take to $0$ the elements of the 2-torsion submodule of $R$ (let's call it $T$). It follows by commutativity that $\mathrm{\ker \partial_*}=T,$ and hence $H_2(\mathbb{R}P^2;R)\simeq T$. We also have that $H_1(\mathbb{R}P^2)=\mathrm{coker \partial_*},$ which is the map $r \mapsto r+r$, hence $H_1(\mathbb{R}P^2)\simeq R/2R$.
*PS: We may not care about specifying generators for the computations regarding the diagram itself, but the orientation is important in the computation of the degree of the third $f_*$. This is due to the fact that it is not a matter of $\pm 1$ as in the other instances, but of $2$ and $0$ instead.
