Construct a meromorphic function has a double order pole at the points $2^n$ and a double order zero at the points $−2^n$

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I am trying to the following problem:

Fix a complex number $0 < |\lambda| < 1$, and choose $\mu$ with $\mu^ 2 = \lambda$. Define $$T(z) =\sum_{n=-\infty}^{\infty} \mu^{n^2} z^n.$$ Show the following properties of $T$:

1. $T$ converges locally uniformly in $\mathbb{C}^{\ast}$;

2. $T(z) = \mu \cdot z \cdot T(\lambda z)$;

3. $T(−\mu) = 0$;

4. $T$ has in the annulus $|\lambda| < |z| < 1$ exactly one zero;

5. Construct a meromorphic function $$ in $\mathbb{C}^{\ast}$ that has a double order pole at the points $2^n$ and a double order zero at the points $−2^n$ for all $n\in \mathbb{Z}$, is holomorphic elsewhere and satisfies $f(2z)=f(z)$ for all $z\in\mathbb{C}^{\ast}$.


My try:

$1,2,3$ is easy. And for $5$, it is natural to consider $$f(z)=\prod_{n\in \mathbb{Z}}\frac{(z+2^n)^2}{(z-2^n)^2}$$ which is obviously divergent everywhere. How to modify this? How to use the former result? Any hint for the others ($4$)? Thanks!