I got help (Thanks to Mako) to show this result:
\begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{1}{2})} = \frac{2}{(q+1)} \end{gathered} \end{equation}
The Gammas can be rewritten as:
$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$
Using $\Gamma(x+1) = x\Gamma(x)$ We can simplify the expressions above by considering $$x = \frac{1}{2}+\frac{1}{q-1}$$ $$y = \frac{1}{q-1}$$
So that we have the following,
$$\frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+1)\Gamma(y)}$$
Now using the identity $\Gamma(x+1) = x\Gamma(x)$, we can write out
$$ = \frac{y\Gamma(x)\Gamma(y)}{x\Gamma(x)\Gamma(y)}$$
and now simplifying by cancelling $\Gamma(x)$ and $\Gamma(y)$ we get
$$ = \frac{y}{x}$$ which when plugging expressions for $x$ and $y$ back in yield
$$ = \frac{1}{q-1}\frac{1}{\frac{1}{2}+\frac{1}{q-1}}$$ $$ = \frac{2(q-1)}{(q-1)(q+1)}$$ $$ = \frac{2}{q+1}$$
However, I´m trying to obtain the following results unsuccessfully:
\begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{3}{2})} = \frac{4(q-1)}{(q+1)(3q-1)} \end{gathered} \end{equation} And: \begin{equation} \begin{gathered} \frac{\Gamma(\frac{1}{2}+\frac{1}{q-1})}{\Gamma(\frac{1}{q-1})}\frac{\Gamma(\frac{q}{q-1})}{ \Gamma(\frac{q}{q-1}+\frac{5}{2})} = \frac{8(q-1^2)}{(q+1)(3q-1)(5q-3)} \end{gathered} \end{equation}
I´ve tried to rewrite the $\frac{3}{2}$ as:
$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$
And the $\frac{5}{2}$ as:
$$\frac{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}\right)\Gamma\left(\frac{1}{q-1}+1\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{q-1}+1+1+1\right)\Gamma\left(\frac{1}{q-1}\right)}$$
Repeating the same steps above for the first Gamma Function , I can use $\Gamma(x+1) = x\Gamma(x)$ and achieve $\frac{y}{x^2}$ and $\frac{y}{x^3}$ respectivelly for these Gamma Functions with $\frac{3}{2}$ and $\frac{5}{2}$ arguments. However this path is not leading me to the desired solutions which were obtainned by MAPLE software.
I'm not entirely sure what mistake you're making. Both reductions should indeed follow immediately from the identity $\Gamma(x+1)=x\,\Gamma(x)$.
For the first one, we have
$$\begin{align*} \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\Gamma\left(\frac q{q-1}\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\Gamma\left(\frac q{q-1} + \frac32\right)}} &= \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\Gamma\left(\frac1{q-1}+1\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\Gamma\left(\frac 1{q-1} + \frac52\right)}} \\ &= \frac{\color{red}{\Gamma\left(\frac1{q-1} + \frac12\right)} \color{blue}{\frac1{q-1} \Gamma\left(\frac1{q-1}\right)}}{\color{blue}{\Gamma\left(\frac1{q-1}\right)} \color{red}{\left(\frac 1{q-1} + \frac32\right) \left(\frac 1{q-1} + \frac12\right) \Gamma\left(\frac 1{q-1} + \frac12\right)}} \\ &= \frac{\frac1{q-1}}{\left(\frac 1{q-1} + \frac32\right) \left(\frac 1{q-1} + \frac12\right)} = \frac{4(q-1)}{(q+1)(3q-1)} \end{align*}$$
and the second is nearly identical except for an additional factor of $\dfrac1{q-1}+\dfrac52$ in the denominator.