Nonzero solutions of an infinite product Let $-\frac{1}{2}\le a \le\frac{1}{2}$ and $b\in[0,\infty)$.
Definitions:  $$f_k(a;b):=\frac{(2k+\frac{1}{2}+a)^2+b}{(2k+\frac{1}{2}-a)^2+b}(\frac{k}{k+1})^{2a},$$
$$f(a;b):=\prod\limits_{k=1}^\infty f_k(a;b)$$
QUESTIONs: 
(1) Does $f(a;b)=1$ have any solution with $a\neq 0$?
(2) If yes: Single points $(a;b)$ or areas ?    
Thank you very much !
EDIT: Have changed $(\frac{k}{k+1})^a$ to $(\frac{k}{k+1})^{2a}$. It was a mistake.
2th EDIT: It seems to be $f(a,b)<\pi^{2a}$ for $a>0$, at least for e.g. $b>2$. Correct ?   
 A: Let us show that for each $a\in[-\frac12,\frac12]\setminus\{0\}$ there is a unique $b\in(0,\infty)$ such that $f(a;b)=0$. 
By factoring $(2k+\frac{1}{2}\pm a)^2+b$ 
as $4 [k + \frac14\, (1 \pm 2 a - 2 i\sqrt b)][k + \frac14\, (1 \pm 2 a + 2 i\sqrt b)]$, 
noting that $\prod_{k=1}^n(k+c)=\Gamma(n+1+c)/\Gamma(1+c)$ for complex $c$, 
and then using the Stirling formula (see e.g. formula (1) in [Kilbas--Saigo]), we have this closed-form expression for $f(a;b)$:
\begin{equation}
 f(a;b)=\frac{\Gamma(\frac54-\frac a2+\frac{i\sqrt b}2)\Gamma(\frac54-\frac a2-\frac{i\sqrt b}2)}
 {\Gamma(\frac54+\frac a2+\frac{i\sqrt b}2)\Gamma(\frac54+\frac a2-\frac{i\sqrt b}2)}
= \bigg(\frac{\big|\Gamma(\frac54-\frac a2+\frac{i\sqrt b}2)\big|}
 {\big|\Gamma(\frac54+\frac a2+\frac{i\sqrt b}2)\big|}\bigg)^2 . \tag{1}
\end{equation}
By formula $(1)$, 
\begin{equation}
 f(a;0)=\Big(\frac{g(a)}{g(-a)}\Big)^2, 
\end{equation}
where $g(a):=\Gamma(\frac54-\frac a2)$ is convex in $a$. Since $g'(-\frac4{10})>0$, it follows that $g$ is increasing on $[-\frac4{10},\frac12]$, and so, $f(a;0)>1$ for $a\in[0,\frac4{10}]$. Also, for $a\in(\frac4{10},\frac12]$ one has $g(a)>g(\frac4{10})>g(-\frac12)\vee g(-\frac4{10})\ge g(-a)$, and so, $f(a;0)>1$ for $a\in(\frac4{10},\frac12]$ as well. Thus, $f(a;0)>1$ for all $a\in(0,\frac12]$, which is equivalent to $f(a;0)<1$ for all $a\in[-\frac12,0)$. 
Next, using again formula $(1)$ above and formula (3) in [Kilbas--Saigo], it is not hard to see that that $f(a;\infty-)=0<1$ for all $a\in(0,\frac12]$ and $f(a;\infty-)=\infty>1$ for all $a\in[-\frac12,0)$. 
Also, $f(a;b)$ is (strictly) increasing in $b\ge0$ for $a\in[-1/2,0)$ and decreasing in $b\ge0$ for $a\in(0,1/2]$ -- because $f_k(a;b)$ has these properties, for each $k$. 
It follows that indeed for each $a\in[-\frac12,\frac12]\setminus\{0\}$ there is a unique $b\in(0,\infty)$ such that $f(a;b)=1$.

For an illustration of the above result, here are graphs of $f(a,b)$ and $\text{sgn}(f(a,b)-1)$, suggesting something like a straight line of roots at $b\approx1.6$. 

The "straight" line is not quite straight, as the following regional plots of $f(a,b)>1$ suggest: 

A: Numerically, I get this ($500$ factors)...
red is ${}> 1$, yellow is ${} < 1$.

the jagged parts are probably just from numerical approximation.  I'm guessing $f=1$ is only curves, and always $b < 1.8$.  
The Gamma function formula yields a similar picture.
Here is the interesting location:

A: Note that for any $a\in[-\frac12,\frac12]\setminus\{0\}$ and $b\in[0,\infty)$ one has $f_k(a;b)-1\sim a/k$ as $k\to\infty$. So, the product diverges to $0$ if $a<0$ and to $\infty$ if $a>0$. Thus, the equation $f(a;b)=1$ has no solutions with $a\neq 0$. 

Details on $f_k(a;b)-1\sim a/k$ as $k\to\infty$: 
\begin{equation}
 f_k(a;b)=\Big(1+\frac{2a(4k+1)}{(2k+\frac12-a)^2+b}\Big)
 \Big(1-\frac1{k+1}\Big)^a
\end{equation}
\begin{equation}
  =\Big(1+\frac{(2+o(1))a}k\Big)\Big(1-\frac{(1+o(1))a}k\Big)=1+\frac{(1+o(1))a}k. 
\end{equation}
