Let's consider some subset of natural numbers $Q$ and product: $$\prod_{k \in Q}{\frac{k^2-1}{k^2+1}}$$
There is well known theorem that if $Q$ is the set of prime numbers than the above product is rational and $$\prod_{p \in P}{\frac{p^2-1}{p^2+1}}=\frac{2}{5}$$ Can be found in: https://mathworld.wolfram.com/PrimeProducts.html
Actually the question is, is there any other known infinite subset of natural numbers which brings us to rational value, except the prime/composite numbers.
Some more examples: If $Q=N_n = \{t n ~ |~n \in N ~\& ~t \in N\}$ then $$\prod_{k \in N_n}{\frac{k^2-1}{k^2+1}} = \sin \left(\frac{\pi }{n}\right) \text{csch}\left(\frac{\pi }{n}\right)$$
and most probably those values are irrational as of $e^\pi$.
If $1\in Q$, then the product is $0$ -- the $k=1$ factor kills it immediately. That's indeed an uninteresting case.
Otherwise, since each term decreases the product, the smallest result we can get is to include all $k\ge 2$. Now that smallest result, $$ \alpha = \prod_{k=2}^{\infty} \frac{k^2-1}{k^2+1}, $$ is positive, because we can take logarithms factor for factor and get $$ \tag{*} -\log \alpha = \sum_{k=2}^{\infty} \log\frac{k^2+1}{k^2-1} = \sum_{k=2}^{\infty} \log\Bigl(1+\frac{2}{k^2-1}\Bigr) < \frac{2\pi^2}{6} $$ because $\frac{1}{k^2-1} < \frac{1}{(k-1)^2} $ when $k\ge 2$, and and $\log(1+x)\le x$.
However, every real number $\beta\in(\alpha,1)$ -- and in particular every rational number in that interval -- can be produced by some $Q$.
All we have to do is to find a subset of the $\log(1+2/(k^2-1))$ terms that sum to $-\log \beta$. This is always possible in this case -- we can simply start from $k=2$ and select every term that doesn't make the sum so far too large.
There's a subtlety, however: This "greedy" procedure doesn't work for all series. For example, we clearly can't pick a subset of the terms in $\sum_{k=1}^{\infty} 10^{-k}$ to reach an arbitrary number between $0$ and $0.111\ldots$.
What saves us in this case is that each term in (*) is at least half the size of the previous term. (This is easily the case asymptotically, and can be verified by direct computation for the first few terms). Thus whenever we skip a term because we're too close to the target, the sum of the terms we haven't considered yet will be larger than the term we skipped. Therefore the difference between the partial sum and our target number remains bounded by the sum of not-yet-considered terms -- and that bound goes to $0$.
Note that the resulting $Q$ is a computable subset of $\mathbb N$, as long as we can effectively compare $\beta$ to the partial products that would result if we pick a candidate term. That's clearly the case when $\beta$ is rational, and also for a large class of interesting irrational numbers. (We don't need to compute with the logarithms during the practical calculation; that was just for proving the procedure gives the right result).
For example, suppose we want to find a $Q$ where the product is exactly $\frac12$. We take the factors one by one:
The next included factor ends up being for $k=210$. Continuing this way, we find $Q=\{2,4,6,36, 210, 44100, \ldots\}$, so $$ \frac12 = \frac{2^2-1}{2^2+1} \times \frac{4^2-1}{4^2+1} \times \frac{6^2-1}{6^2+1} \times \frac{36^2-1}{36^2+1} \times \frac{210^2-1}{210^2+1} \times \frac{44100^2-1}{44100^2+1} \times \cdots $$ This is by far not the only $Q$ that produces $\frac12$, though.