The question
Consider the representation: $$ \frac{π^2}{\sin^2(πz)} = \sum_{n∈ℤ}\frac{1}{(z+n)^2} \tag{0}$$ valid for all $z ∈ ℂ \setminus ℤ$. Deduce that the Taylor series of $z^2/\sin^2(z)$ for $|z|<π$ is: $$ \frac{z^2}{\sin^2{z}} = 1 + 2\sum_{m=1}^∞(2m-1) \frac{ζ(2m)}{π^{2m}}z^{2m} \tag{0.1}$$
Deduce also that: $$ ζ(2) = \sum_{n=1}\frac{1}{n^2} = \frac{π^2}{6}, \qquad ζ(4) = \sum_{n=1}^∞\frac{1}{n^4} = \frac{π^4}{90} $$
Finally explain why for all $m ≥ 1$, we have: $$ \frac{ζ(2m)}{π^{2m}} = \frac{1}{π^{2m}}\sum_{n=1}^∞\frac{1}{n^{2m}} $$
What I've tried
For the first and third question I have no idea of how to start. And for the second question I've been able to see that because of the sine infinite products expansion one can have:
$$ \sin(πz) = πz\prod_{n=1}^∞\left(1 - \frac{z^2}{n^2}\right) $$ $$ πz - \frac{π^3}{3!} + … = πz \left( 1 - \sum_{n=1}^∞\frac{z^2}{n^2} + … \right) $$ And then: $$ \frac{π^3}{3!} = π\sum_{n=1}^∞\frac{1}{n^2} $$ so necessarily we have $\frac{π^2}{6} = \sum_{n=1}^∞\frac{1}{n^2}$. But that calculus does not come from the expression given at the beginning of the question.
Any ideas for the 3 questions?
[Edit] - based on what Daniel Fischer♦ suggested
If we want to expand $\frac{z^2}{(z+n)^2}$ into a series, calculating its successive derivatives we get: $$\left( \frac{z^2}{(z+n)^2} \right)^{(j)}\Big|_{z=0} = (-1)^j\frac{(j-1)n^2}{n^j}\tag{1},\quad \text{with}\quad j>1$$ and $0$ when $j=0,1$.
Then one can express the fraction as follows: $$ \frac{z^2}{(z+n)^2} = \sum_{j=2}^∞(-1)^j\frac{(j-1)}{n^j}z^j \tag{2}$$ And summing over $n$ gives: $$ \sum_{j=2}^∞(-1)^j(j-1)z^j \sum_{n∈ℤ} \frac{1}{n^j} \tag{3}$$ If $j$ is odd the sum vanishes, and when it's even we get twice $ζ(2j)$. So $(3)$ equals to: $$ 2·\sum_{j=1}^∞(2j-1)z^{2j}ζ(2j) $$
Recalling $(0)$ we have: $$ \frac{(πz)^2}{\sin^2(πz)} = 2·\sum_{j=1}^∞(2j-1)z^{2j}ζ(2j) $$ so: $$ \frac{z}{\sin^2(z)} = 2·\sum_{j=1}^∞(2j-1)\frac{ζ(2j)}{π^{2j}}z^{2j} \tag{4}$$
But $(4)$ differs from $(0.1)$ in a unity. What am I doing wrong?
Given: $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ we have: $$ \log\sin(\pi z)-\log(\pi z) = \sum_{n\geq 1}\log\left(1-\frac{z^2}{n^2}\right)\tag{2}$$ and by differentiating both sides twice we get: $$ -\frac{\pi^2}{\sin^2(\pi z)}+\frac{1}{z^2}=-\sum_{n\geq 1}\left[-\frac{1}{(z-n)^2}-\frac{1}{(z+n)^2}\right]\tag{3}$$ or: $$ \frac{\pi^2}{\sin^2(\pi z)}=\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}\tag{4} $$ where both sides are meromorphic functions with double poles at the integers, with residue zero and singular part behaving like $\frac{1}{z^2}$. So: $$ \frac{\pi^2 z^2}{\sin^2(\pi z)} = 1+\sum_{n\neq 0}\sum_{m\geq 1}(2m-1)n^{-2m}x^{2m}=1+2\sum_{m\geq 1}(2m-1)\zeta(2m) x^{2m}\tag{5} $$ where both sides are regular functions in a neighbourhood of zero. The values of $\zeta(2m)$ can so be computed through the Taylor series of the LHS.