Are these 5 the only eta quotients that parameterize $x^2+y^2 = 1$?

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.

Courtesy of Somos' answer, we can find more eta quotient parameterizations to $x^2+y^2 = 1$. Define,

$$a(\tau) = \frac{\eta^3(\tau)\,\eta(6\tau)}{2\,\eta^3(2\tau)\,\eta(3\tau)},\quad\quad b(\tau) = \frac{\eta^2(\tau)\,\eta(6\tau)}{\sqrt3\,\eta^2(3\tau)\,\eta(2\tau)}$$

$$c(\tau) = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)},\quad\quad d(\tau) = \frac{\eta(\tau)\,\eta^2(6\tau)}{\eta(3\tau)\,\eta^2(2\tau)}\quad$$

Note that $c(\tau)$ is the cubic continued fraction and the four obey,

$$\frac{-1}{a^3(\tau)} +\frac1{b^4(\tau)}= 1,\quad\quad \frac{-1}{c^3(\tau)} +\frac1{d^4(\tau)}= 1$$

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For the first identity, recall $\alpha(\tau)$ and $\beta(\tau)$ from the original post. Then we have,

$$\big(\alpha(\tau)\,\alpha(3\tau)\big)^2+\big(\beta(\tau)\,\beta(3\tau)\big)^2=1\tag1\quad$$

as well as,

$$\left(\frac{a^2(2\tau)\,b(\tau)}{a(\tau)\,b^2(2\tau)}\right)^2+\big(2a(\tau)\,a(2\tau)\big)^2=1\quad\tag2$$

$$\left(\frac{c^2(\tau)\,d(2\tau)}{c(2\tau)\,d^2(\tau)}\right)^2+\big(2c(\tau)\,c(2\tau)\big)^2=1\quad\tag3$$

$$\quad\left(\frac{a^2(2\tau)}{a(\tau)}\right)^2-\big(2a(\tau)\,a(2\tau)\big)^2=\big(\sqrt3\,d(2\tau)\big)^2\tag4$$

$$\quad\left(\frac{c^2(\tau)}{c(2\tau)}\right)^2-\big(2c(\tau)\,c(2\tau)\big)^2=\big(\sqrt3\,b(\tau)\big)^2\tag5$$

$$\left(\frac{a^2(2\tau)}{a(\tau)}\right)^2+\left(\frac{2a(\tau)\,a(2\tau)}{b(\tau)}\right)^2=1\quad\tag6$$

$$\left(\frac{c^2(\tau)}{c(2\tau)}\right)^2+\left(\frac{2c(\tau)\,c(2\tau)}{d(2\tau)}\right)^2=1\quad\tag7$$

The first one being $t_{12,12,48b}$ while the rest are the 6 mentioned by Somos.

P.S. This is a revised version of the original answer since, as I suspected, they could be expressed in a more consistent form.