It is a well-known fact that projective line has genus $0$. However, I cannot deduce this fact from Riemann-Roch theorem and Serre duality. Let's recall these results
Let $X$ be smooth irreducible projective curve, $g$ be the genus of $X$ and $D$ a divisor on $X$ then
Riemann-Roch theorem: $$\dim H^0(X,D)-\dim H^1(X,D)=1-g+\deg(D)$$
Combining with Serre duality we have $$\dim H^0(X,D)-\dim H^0(X,K_X-D)=1-g+\deg(D)$$ where $K_X$ is the canonical divisor. We also know that $g=H^1(X,\mathcal{O}_X)=H^0(X,\Omega_X^1)$. Let $D=K_X$ and $D=0$ respectively, we obtain degree of canonical divisor is $2g-2$ and of principal divisor is $0$.
In case of $\mathbb{P}^1$ every divisor $D$ is linear equivalent to $n \infty$ for some $n$. My idea is choosing appropriate divisor and using the formulas above to compute $g= H^1(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1})$ but maybe it doesn't work.
Probably the easiest way to do this is using the Euler exact sequence, which shows that the canonical divisor $K_{\mathbb{P}^1}$ is equivalent to $-2\infty$, which gives you that $g=0$ by Riemann-Roch and Serre duality.
You can also calculate whatever homology groups you're interested in directly using the Cech complex, but that is more laborious.