computing $A_2=\sum_{k=1}^{n}\frac{1}{(z_k-1)^2} $ and $\sum_{k=1}^n \cot^2\left( \frac{k\pi}{n+1}\right)$

That is the core of Cauchy's proof of $\zeta(2)=\frac{\pi^2}{6}$. The whole proof can be found on Wikipedia.

$$ \sum_{k=1}^{n}\cot^2\left(\frac{\pi k}{n+1}\right) = \color{blue}{\frac{n(n-1)}{3}}.$$

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An insane overkill, just for fun. Due to the reflection formula for the trigamma function, $$ \psi'(s)+\psi'(1-s) = \frac{\pi^2}{\sin^2(\pi s)} $$ we have: $$ \sum_{k=1}^{n}\cot^2\left(\frac{\pi k}{n+1}\right)=-n+\frac{2}{\pi^2}\sum_{k=1}^{n}\psi'\left(\frac{k}{n+1}\right)=-n+\frac{2}{\pi^2}\sum_{k=1}^{n}\sum_{m\geq 0}\frac{1}{\left(m+\frac{k}{n+1}\right)^2} $$ where the RHS can be rearranged as $$ -n+\frac{2(n+1)^2}{\pi^2}\!\!\!\sum_{\substack{M\geq 1 \\ (n+1)\nmid M}}\frac{1}{M^2}=-n+\frac{2(n+1)^2}{\pi^2}\zeta(2)\left(1-\frac{1}{(n+1)^2}\right)=\color{blue}{\frac{n(n-1)}{3}}. $$

With $A_2$, it is easy to show that if $$P(z)=\prod\limits_{k=1}^{n}\left(z-z_k\right)$$ then $$P'(z)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(z-z_k\right)$$ and $$\frac{P'(z)}{P(z)}=\sum\limits_{i=1}^{n}\frac{1}{z-z_i} \Rightarrow -\frac{P'(z)}{P(z)}=\sum\limits_{i=1}^{n}\frac{1}{z_i-z} \tag{1}$$ and finally $$A_1=\sum\limits_{i=1}^{n}\frac{1}{z_i-1}=-\frac{P'(1)}{P(1)}=-\frac{n}{2}$$ However, from $(1)$: $$\left(-\frac{P'(z)}{P(z)}\right)^{'}=\left(\sum\limits_{i=1}^{n}\frac{1}{z_i-z}\right)^{'}=\sum\limits_{i=1}^{n}\frac{1}{(z_i-z)^2}$$ and $$\sum\limits_{i=1}^{n}\frac{1}{(z_i-z)^2} = -\frac{P''(z)P(z)-P'(z)^2}{P(z)^2} \tag{2}$$ and $$A_2=-\frac{P''(1)P(1)-P'(1)^2}{P(1)^2}$$ where $$P''(1)=n(n-1)+(n-1)(n-2)+...+3\cdot2+2\cdot1=\frac{(n-1)n(n+1)}{3}$$ thus $$A_2=-\frac{\frac{(n-1)n(n+1)}{3}(n+1)-\left(\frac{n(n+1)}{2}\right)^2}{(n+1)^2}=\frac{4n-n^2}{12}$$