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Question $2$ has a surprisingly simple answer:

If $$\cosh\alpha_j+\cos\frac{\pi j}{2n}=\cosh\beta_k+\cos\frac{\pi k}{2m}=x\tag{3}$$ for all $1\le j\le 2n-1,\ 1\le k\le 2m-1$, then $$ \prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\prod_{k=1}^{2m-1}\left(\frac{\tanh n\beta_k}{\sinh\beta_k}\right)^{(-1)^k}. \tag{4} $$ Proof. We use two well known formulas $$ 2^{m-1} \prod _{j=1}^{m-1} \left(\cosh \alpha-\cos \frac{\pi j}{m}\right)=\frac{\sinh m \alpha}{\sinh \alpha}, $$ $$ 2^{m-1} \prod _{j=1}^m \left(\cosh \alpha-\cos \frac{\pi (j-1/2)}{m}\right)=\cosh m \alpha, $$ to write the products in $(4)$ in symmetric form: \begin{align} &\prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\tanh m\alpha_{2j}}{\sinh\alpha_{2j}}\frac{\sinh\alpha_{2j-1}}{\tanh m\alpha_{2j-1}}\\ &=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (m-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(m-1/2)}{m}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (k-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(k-1/2)}{m}}\frac{\cosh\alpha_{2j}-\cos\frac{\pi k}{m}}{\cosh\alpha_{2j-1}-\cos\frac{\pi k}{m}}\\ &=\frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi (k-1/2)}{m}}{x-\cos\frac{\pi j}{n}-\cos\frac{\pi(k-1/2)}{m}}\frac{x-\cos\frac{\pi j}{n}-\cos\frac{\pi k}{m}}{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi k}{m}}. \end{align} Now, in this last formula the factor $$ \frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}=\frac{1}{x+\cos\frac{\pi}{2n}+\cos\frac{\pi }{2m}} $$ is obviously symmetric when $m$ and $n$ are interchanged, and so is the factor $\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}$ and the double product. So both sides in $(4)$ are symmetric when $m$ and $n$ are interchanged, and hence they are equal. $~\Box$

EDIT (Feb 2024): The question has been generally answered in the article:

• Martin Nicholson, Finite and infinite product transformations, arXiv:1712.06097.

Question $2$ has a surprisingly simple answer:

If $$\cosh\alpha_j+\cos\frac{\pi j}{2n}=\cosh\beta_k+\cos\frac{\pi k}{2m}=x\tag{3}$$ for all $1\le j\le 2n-1,\ 1\le k\le 2m-1$, then $$ \prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\prod_{k=1}^{2m-1}\left(\frac{\tanh n\beta_k}{\sinh\beta_k}\right)^{(-1)^k}. \tag{4} $$ Proof. We use two well known formulas $$ 2^{m-1} \prod _{j=1}^{m-1} \left(\cosh \alpha-\cos \frac{\pi j}{m}\right)=\frac{\sinh m \alpha}{\sinh \alpha}, $$ $$ 2^{m-1} \prod _{j=1}^m \left(\cosh \alpha-\cos \frac{\pi (j-1/2)}{m}\right)=\cosh m \alpha, $$ to write the products in $(4)$ in symmetric form: \begin{align} &\prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\tanh m\alpha_{2j}}{\sinh\alpha_{2j}}\frac{\sinh\alpha_{2j-1}}{\tanh m\alpha_{2j-1}}\\ &=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (m-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(m-1/2)}{m}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (k-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(k-1/2)}{m}}\frac{\cosh\alpha_{2j}-\cos\frac{\pi k}{m}}{\cosh\alpha_{2j-1}-\cos\frac{\pi k}{m}}\\ &=\frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi (k-1/2)}{m}}{x-\cos\frac{\pi j}{n}-\cos\frac{\pi(k-1/2)}{m}}\frac{x-\cos\frac{\pi j}{n}-\cos\frac{\pi k}{m}}{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi k}{m}}. \end{align} Now, in this last formula the factor $$ \frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}=\frac{1}{x+\cos\frac{\pi}{2n}+\cos\frac{\pi }{2m}} $$ is obviously symmetric when $m$ and $n$ are interchanged, and so is the factor $\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}$ and the double product. So both sides in $(4)$ are symmetric when $m$ and $n$ are interchanged, and hence they are equal. $~\Box$

The question has been generally answered in this article https://arxiv.org/abs/1712.06097

Question $2$ has a surprisingly simple answer:

If $$\cosh\alpha_j+\cos\frac{\pi j}{2n}=\cosh\beta_k+\cos\frac{\pi k}{2m}=x\tag{3}$$ for all $1\le j\le 2n-1,\ 1\le k\le 2m-1$, then $$ \prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\prod_{k=1}^{2m-1}\left(\frac{\tanh n\beta_k}{\sinh\beta_k}\right)^{(-1)^k}. \tag{4} $$ Proof. We use two well known formulas $$ 2^{m-1} \prod _{j=1}^{m-1} \left(\cosh \alpha-\cos \frac{\pi j}{m}\right)=\frac{\sinh m \alpha}{\sinh \alpha}, $$ $$ 2^{m-1} \prod _{j=1}^m \left(\cosh \alpha-\cos \frac{\pi (j-1/2)}{m}\right)=\cosh m \alpha, $$ to write the products in $(4)$ in symmetric form: \begin{align} &\prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\tanh m\alpha_{2j}}{\sinh\alpha_{2j}}\frac{\sinh\alpha_{2j-1}}{\tanh m\alpha_{2j-1}}\\ &=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (m-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(m-1/2)}{m}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (k-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(k-1/2)}{m}}\frac{\cosh\alpha_{2j}-\cos\frac{\pi k}{m}}{\cosh\alpha_{2j-1}-\cos\frac{\pi k}{m}}\\ &=\frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi (k-1/2)}{m}}{x-\cos\frac{\pi j}{n}-\cos\frac{\pi(k-1/2)}{m}}\frac{x-\cos\frac{\pi j}{n}-\cos\frac{\pi k}{m}}{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi k}{m}}. \end{align} Now, in this last formula the factor $$ \frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}=\frac{1}{x+\cos\frac{\pi}{2n}+\cos\frac{\pi }{2m}} $$ is obviously symmetric when $m$ and $n$ are interchanged, and so is the factor $\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}$ and the double product. So both sides in $(4)$ are symmetric when $m$ and $n$ are interchanged, and hence they are equal. $~\Box$

Question $2$ has a surprisingly simple answer:

If $$\cosh\alpha_j+\cos\frac{\pi j}{2n}=\cosh\beta_k+\cos\frac{\pi k}{2m}=x\tag{3}$$ for all $1\le j\le 2n-1,\ 1\le k\le 2m-1$, then $$ \prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\prod_{k=1}^{2m-1}\left(\frac{\tanh n\beta_k}{\sinh\beta_k}\right)^{(-1)^k}. \tag{4} $$ Proof. We use two well known formulas $$ 2^{m-1} \prod _{j=1}^{m-1} \left(\cosh \alpha-\cos \frac{\pi j}{m}\right)=\frac{\sinh m \alpha}{\sinh \alpha}, $$ $$ 2^{m-1} \prod _{j=1}^m \left(\cosh \alpha-\cos \frac{\pi (j-1/2)}{m}\right)=\cosh m \alpha, $$ to write the products in $(4)$ in symmetric form: \begin{align} &\prod_{j=1}^{2n-1}\left(\frac{\tanh m\alpha_j}{\sinh\alpha_j}\right)^{(-1)^j}=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\tanh m\alpha_{2j}}{\sinh\alpha_{2j}}\frac{\sinh\alpha_{2j-1}}{\tanh m\alpha_{2j-1}}\\ &=\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\prod_{j=1}^{n-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (m-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(m-1/2)}{m}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{\cosh\alpha_{2j-1}-\cos\frac{\pi (k-1/2)}{m}}{\cosh\alpha_{2j}-\cos\frac{\pi(k-1/2)}{m}}\frac{\cosh\alpha_{2j}-\cos\frac{\pi k}{m}}{\cosh\alpha_{2j-1}-\cos\frac{\pi k}{m}}\\ &=\frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}\\ &\times \prod_{j=1}^{n-1}\prod_{k=1}^{m-1}\frac{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi (k-1/2)}{m}}{x-\cos\frac{\pi j}{n}-\cos\frac{\pi(k-1/2)}{m}}\frac{x-\cos\frac{\pi j}{n}-\cos\frac{\pi k}{m}}{x-\cos\frac{\pi (j-1/2)}{n}-\cos\frac{\pi k}{m}}. \end{align} Now, in this last formula the factor $$ \frac{1}{\cosh\alpha_{2n-1}-\cos\frac{\pi (m-1/2)}{m}}=\frac{1}{x+\cos\frac{\pi}{2n}+\cos\frac{\pi }{2m}} $$ is obviously symmetric when $m$ and $n$ are interchanged, and so is the factor $\frac{\sinh\alpha_{2n-1}}{\tanh m\alpha_{2n-1}}\frac{\sinh\beta_{2m-1}}{\tanh n\beta_{2m-1}}$ and the double product. So both sides in $(4)$ are symmetric when $m$ and $n$ are interchanged, and hence they are equal. $~\Box$

The question has been generally answered in this article https://arxiv.org/abs/1712.06097