First consider the homology of the pair $(X^i \cup A^{i+1}, X^{i-1} \cup A^i)$. This is a CW pair, so by excision $$H_i(X^i \cup A^{i+1}, X^{i-1} \cup A^i) \cong \tilde{H}_i(X^i \cup A^{i+1} / X^{i-1} \cup A^i) \cong \tilde{H}_i (X^i / X^{i-1} \cup A^i) \cong H_i(X^i, X^{i-1} \cup A^i)$$where the second isomorphism holds because adding the $(i+1)$-cells of $A$ will have no effect on $H_i$ since we're collapsing the $i$-skeleton of $A$ to a point in the quotient (this is basically a relative version of Lemma 2.34 in your book).
This means that we can substitute this into the long exact sequence of the pair $(X^i \cup A^{i+1}, X^{i-1} \cup A^i)$ to get
$$\cdots \to H_i(X^{i-1} \cup A^i) \to H_i(X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1} \cup A^i) \to H_{i-1}(X^{i-1} \cup A^i) \to \cdots.$$
But $X^i \cup A^{i+1}$ fits into another long exact sequence, too: the long exact sequence of the pair $(X^{i+1} \cup A^{i+2}, X^i \cup A^{i+1})$, which is given by
$$\cdots \to H_{i+1}(X^{i+1} \cup A^{i+2}) \to H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i \cup A^{i+1}) \to \cdots$$
where we've identified the relative group in the middle by the same isomorphism above with $i$ replaced by $i+1$.
so define the map $H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1}\cup A^i)$ to be the composition $H_{i+1}(X^{i+1}, X^i \cup A^{i+1}) \to H_i(X^i \cup A^{i+1}) \to H_i(X^i, X^{i-1} \cup A^i)$, where the first map is from the second long exact sequence above, and the second map is from the first long exact sequence above.
The remaining steps are to show that the sequence of these maps forms a chain complex and that the homology of this chain complex yields the homology of the pair $(X,A)$.
Let $X,Y$ be nonempty path-connected spaces, $\alpha \in H^p(X;k), \beta \in H^q(Y;k)$ for some field $k$.
Then you get cohomology classes $\alpha \otimes 1 \in H^p(X;k)\otimes H^0(Y;k) \subset H^p(X\times Y;k)$ and $1\otimes \beta \in H^0(X;k)\otimes H^q(Y;k) \subset H^q(X\times Y;k)$.
I claim that the inclusion $i:X\to X\times Y$ at any basepoint of $Y$ sends $\alpha\otimes 1$ to $\alpha$ and $1\otimes \beta$ to $0$. If you apply this in your particular situation with $X=Y= \mathbb RP^{n-1}$ , $\alpha_1$ and $\alpha_2$, you get the desired result.
To prove the claim, you can use the projection $p:X\times Y \to X$ and notice that $p\circ i = id_X$, so that when you look at cohomology, $i^*\circ p^* = id$ on cohomology. In particlar, $i^*(\alpha\otimes 1) = i^*(p^*(\alpha)) = \alpha$. Here, I used $p^*\alpha = \alpha\otimes 1$, but this is essentially part of the Künneth theorem.
Similarly, we may use the projection $q: X\times Y\to Y$ and notice that $q\circ i$ is a constant map, hence it induces $0$ on cohomology. It follows that $0= i^*(q^*(\beta)) = i^*(1\otimes \beta)$; using $q^*\beta = 1\otimes \beta$, which again is essentially part of the Künneth theorem.
This solces the question of "why does the inclusion act like that on $\alpha_1, \alpha_2$ ?"
Then for the second question (how the next sentence follows), consider then $i^*h^*\alpha = i^*(k_1\alpha_1+k_2\alpha_2 ) = k_1i^*\alpha_1 + k_2i^*\alpha_2 = k_1\alpha$ by the previous computation, and $h\circ i$ is a homeomorphism (by what has been said before).
Therefore $i^*h^* = (h\circ i)^*$ is an isomorphism on cohomology, and since $\alpha\neq 0$, it must be that $i^*h^*\alpha \neq 0$, i.e. $k_1\neq 0$.
Best Answer
$1)$ For a continuous function $f\colon X \to Y$ the long exact sequence for the mapping cone comes from the long exact sequence for the pair $(M_f, X)$, where the mapping cylinder $M_f$ is homotopy equivalent to $Y$ and $C_f = M_f / X$. Since $(M_f, X)$ is a "good" pair the quotient map induces an isomorphism of homology groups $H_n(M_f, X; G) \cong \tilde{H}_n(C_f; G)$ for any $G$.
$2)$ If $f$ induces isomorphisms $H_n(X;G)\cong H_n(Y;G)$ for each $n$, then by exactness of the mapping cone sequence $\tilde{H}_n(C_f;G) \cong 0$ for all $n$. In fact the converse is also true: if $\tilde{H}_n(C_f;G) \cong 0$ for all $n$ then $f$ is an $H_*(-;G)$ isomorphism.
With these facts you should be able to use part $(a)$ to deduce part $(b)$.