(This is an extended comment, not a true answer. It provides a sort-of-closed-form expression for $f_+(n)$.)
There is a special function $S_2(\alpha; z)$, called the double sine function, which is meromorphic in $z \in \mathbb{C}$ and which satisfies $$ S_2(\alpha; z + 1) = \frac{S_2(\alpha; z)}{2 \sin(\tfrac{\pi}{\alpha} z)} \qquad \text{and} \qquad S_2(\alpha; z + \alpha) = \frac{S_2(\alpha; z)}{2 \sin(\pi z)} \, . $$ If we set $\alpha = 2 \pi$ and write simply $S_2(z) = S_2(2 \pi, z)$, we get $$ S_2(z + 1) = \frac{S_2(z)}{2 \sin(\tfrac{z}{2})} \, . $$ Choose $b > 0$ such that $\cosh \xi = \tfrac{5}{4}$, and write $\xi_\pm = \tfrac{\pi}{4} \pm b i$. Then $$ \sin(z) + \tfrac{5}{4} = 2 \sin(\tfrac{z}{2} + \xi_+) \sin(\tfrac{z}{2} + \xi_-) . $$ It follows that $$ \frac{2^n S_2(\xi_+) S_2(\xi_-)}{S_2(n + 1 + \xi_+) S_2(n + 1 + \xi_-)} = \prod_{k = 0}^n 2 \sin(\tfrac{k}{2} + \xi_+) \sin(\tfrac{k}{2} + \xi_-) = \prod_{k = 0}^n (\sin k + \tfrac{5}{4}) $$ is your sequence $f_+(n)$.
Now the double sine function is a special function that I know next to nothing about (I learned about it, as well as about many other fancy special functions, from Alexey Kuznetsov), but apparently it is reasonably well-understood. The refernences that I got from Alexey are:
S. Koyama and N. Kurokawa, Multiple sine functions, Forum Mathematicum 15 (2006), no. 6, 839–876;
S. Koyama and N. Kurokawa, Values of the double sine function, J. Number Theory 123 (2007), no. 1, 204–223.
I did not check them to see if they contain any information relevant to your questions.