Homology group of Real projective plane
You'll want to use the fact that $\mathbb{R}P^n$ can be written as $\mathbb{R}P^{n-1}\cup_f D^n$ where $D^n$ is the $n$-dimensional ball, and $f\colon S^{n-1}\to\mathbb{R}P^{n-1}$ is a 2-fold covering map, so we are gluing the $n$-ball along its boundary to $\mathbb{R}P^{n-1}$ via this map. You can then use Mayer-Vietoris and induced maps to explicitly work out the connecting map. In your case, you have $\mathbb{R}P^2=M\cup_f D^2$, where $M$ is the Mobius strip and $f\colon S^1\to M$ is the doubling map up to homotopy, or just the inclusion of the boundary into the Mobius strip.
To be more explicit, via Mayer-Vietoris, we get a long exact sequence $$\cdots\to H_2(M)\oplus H_2(D^2)\to H_2(\mathbb{R}P^2)\to H_1(S^1)\to H_1(M)\oplus H_1(D^2)\to\cdots$$ which, using the fact that $H_2(M)=H_2(D^2)=H_1(D^2)=0$ and $H_1(S^1) \cong H_1(M) \cong\mathbb{Z}$, reduces to the exact sequence. $$\cdots\to 0\to H_2(\mathbb{R}P^2) \stackrel{g}{\to} \mathbb{Z} \stackrel{\times 2}{\to} \mathbb{Z} \to\cdots$$
where we get that $\times 2$ map in the above sequence from the fact that the inclusion of the intersection of the two spaces (homotopy equivalent to a circle) into the Mobius strip is (up to homotopy) the degree-$2$ covering map, which induces multiplcation by $2$ in first homology.
By exactness, the image of $g$ must be $0\subset\mathbb{Z}$ as the doubling map in injective, but $g$ must itself be injective by exactness because the map $0\to H_2(\mathbb{R}P^2)$ has trivial image. The only way both of these conditions on $g$ can be satisfied is if $H_2(\mathbb{R}P^2)$ is trivial.
The real projective plane is the space of all lines in $\mathbb R^3$. One can also treat it as $\mathbb S^2/\sim$, where $x\sim -x$. So, it can also be treated as the upper hemisphere together with the equator with the identification $z\sim z^2$ on the equator. So the gluing map $\delta_2$ is of degree two.