This question follows a previous question Sheaf morphism from closed subscheme is a closed immersion, it's just another part so I'll recall everything.
For $K=\bar{K}$ a field consider $X=\mathbb P^1_K$, $Z=\{P_1,\dots,P_n\}\subseteq X$ closed points. Give $Z$ the reduced induced closed subscheme structure and write $\iota:Z\to X$ the closed immersion.
I've shown that $\ker \iota^{\#}$ is an invertible $\mathcal O_X$-module isomorphic to $\mathcal O_X(D)^{-1}$ where $D$ is the Cartier divisor corresponding to $Z$.
Now I aim to find $\dim_K(H^i(X,\ker \iota^{\#}))$ for $i=0,1$. These are defined by considering an injective resolution $I^{\bullet}$ of $\ker \iota^{\#}$ and computing $H^i(\Gamma(X,I^{\bullet}))$, and in fact we can only take a flasque resolution. We have and exact sequence $0\to \ker \iota^{\#}\to \mathcal O_X \to \iota_{*}\mathcal O_Z\to 0$ which I think will be the resolution we need.
Now applying $\Gamma(X,-)$ we get $0\to \Gamma(X,\ker \iota^{\#})(=0?)\to K\to K^n\to 0$ where $\Gamma(X,\iota_{*}\mathcal O_Z)=\mathcal O_Z(Z)=K^n$ according to the question I mentioned at the beginning. So we would get $H^0(X,\ker \iota^{\#})=0=\ker f$ where $f:K\to K^n$ is the inclusion in a coordinate (I don't really know what $f$ does, it's just natural) and $H^1(X,\ker \iota^{\#})=K^n/K\cong K^{n-1}$.
My questions are : Is it correct ? How can I show that $i_{*}\mathcal O_Z$ is indeed flasque ?
$0\to \ker\iota^\sharp\to\mathcal{O}_X\to\iota_*\mathcal{O}_Z\to 0$ is not a flasque resolution, since $\mathcal{O}_X$ is not flasque (it has global sections $K$, and sections $K[\frac{x_1}{x_0}]$ over $D_+(x_0)$). It still does suffice to solve your problem, though: from the short exact sequence of sheaves, we get a long exact sequence of cohomology groups, and we note that $\mathcal{O}_X$ has $H^0=K$ and $H^i=0$ for all $i\neq 0$, so $H^0(\ker\iota^\sharp)$ and $H^1(\ker\iota^\sharp)$ are respectively the kernel and cokernel of $H^0(\mathcal{O}_X)\to H^0(\iota_*\mathcal{O}_Z)$. Now we just note that this map is restriction of global sections, so sends $1\in K$ to $(1,1,\cdots,1)\in K^n$, and this gives that $H^0(\ker\iota^\sharp)=0$ and $H^1(\ker\iota^\sharp)\cong K^{n-1}$.
It is also true that $\iota_*\mathcal{O}_Z$ is flasque, though. Since $(\iota_*\mathcal{O}_Z)(U)=\mathcal{O}_Z(\iota^{-1}(U))$, it suffices to show that $\mathcal{O}_Z$ is flasque, and this follows from the fact that $Z$ is discrete: any sheaf $\mathcal{F}$ on a discrete space $S$ has $\mathcal{F}(S_0) = \prod_{s\in S_0} \mathcal{F}_s$ and the restrictions are the obvious projections.