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For your question $2$ all that is needed is a quick search of 'eta07.gp' for three terms identities with even rank and with all even exponents. I turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.

Note that given an eta identity, replacing $q$ with $-q$ gives an eta identity. Also the same with replacing $q^d$ with $q^{N/d}.$ By iterating these to transforms we get an eta identity cluster. There can be up to $12$ identities in a cluster.

For your question $2$ all that is needed is a quick search of 'eta07.gp' for three terms identities with even rank and with all even exponents. I turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.

Note the two transforms. First, given an eta identity, replacing $q$ with $-q$ gives an eta identity. Second, the same with replacing $q^d$ with $q^{N/d}$ where $N$ is the level. By iterating these two transforms we get an eta identity cluster. There can be up to $12$ identities in a cluster.

For your question $2$ a quick search of 'eta07.gp' turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.

For your question $2$ all that is needed is a quick search of 'eta07.gp' for three terms identities with even rank and with all even exponents. I turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.

Note that given an eta identity, replacing $q$ with $-q$ gives an eta identity. Also the same with replacing $q^d$ with $q^{N/d}.$ By iterating these to transforms we get an eta identity cluster. There can be up to $12$ identities in a cluster.

For your question $2$ a quick search of 'eta07.gp' turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ (\sqrt{1/4}u_1^3u_{12}/(u_3u_4^3))^2 + (\sqrt{3/4}u_2^4u_6^2/(u_1u_3u_4^4))^2 = 1 $$ and similarly for the other $5$ identities. Probably other similar clusters can be found.

For your question $2$ a quick search of 'eta07.gp' turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.