Let $F$ and $G$ be elements of the Selberg class with respective gamma factors $\gamma_{F}(s)=Q_{F}^s\prod_{i=1}^{d}\Gamma(s/2+\mu_{i}(F))$ and $\gamma_{G}(s)=Q_{G}^s\prod_{j=1}^{d'}\Gamma(s/2+\mu_{j}(G)) $ .
Are there known examples of $ F $ and $ G $ such that if $ \mu_{i}(F) $ is real for all $ i $ and $ \mu_{j}(G) $ is real for all $ j $ then $ \gamma_{F\otimes G}(s)=Q_{F\otimes G}^s\prod_{i=1}^{d}\prod_{j=1}^{d'}\Gamma(s/2+\mu_{i}(F)\mu_{j}(G)) $ assuming $F\otimes G(s)=\sum_{n>0}\frac{a_{n}(F)a_{n}(G)}{n^s} $ for $ \Re(s)>1 $ whenever $H(s)=\sum_{n>0}\frac{a_{n}(H)}{n^s} $ for $ \Re(s)>1 $ belongs to the Selberg class ?
First off, the whole point of the Selberg class is the following conjecture:
This is known to hold for $n < 2$ (remember that in the Selberg class, $n$ need not be an integer, though we conjecture that there are no elements of the Selberg class of nonintegral $n$). In general, it is known that $L$-functions of automorphic representations are in the Selberg class, but not the converse.
In any case, please, for the love of God, learn about automorphic forms if you want to be able to say anything about properties of $L$-functions. There is no point studying the Selberg class if you don't know what's happening for automorphic forms!
The reason I've gone on this rant is because you continually ask questions like this that can be easily answered if you have read anything about automorphic forms. It seems very backwards to me to ask all these questions about elements of the Selberg class and not understand anything about the $L$-functions that we actually know to be in this class!
Anyway, back to your question. There is a major mistake: the Rankin-Selberg $L$-function $L(s, \pi \otimes \pi')$ of two elements $\pi,\pi'$ in the Selberg class is not equal to $\sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}$ (where of course $L(s,\pi) = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n)}{n^s}$ and $L(s,\pi') = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi'}(n)}{n^s}$). Rather, the correct way to define this is as follows. Write \[L(s,\pi) = \prod_p \prod_{j = 1}^{n} \frac{1}{1 - \alpha_{\pi,j}(p) p^{-s}}, \qquad L(s,\pi) = \prod_p \prod_{k = 1}^{n'} \frac{1}{1 - \alpha_{\pi',j}(p) p^{-s}}.\] Then the correct Rankin-Selberg $L$-function is \[L(s,\pi \otimes \pi') = \prod_{p \nmid (Q_{\pi}, Q_{\pi'})} \prod_{j = 1}^{n} \prod_{k = 1}^{n'} \frac{1}{1 - \alpha_{\pi,j}(p) \alpha_{\pi',k} p^{-s}} \prod_{p \mid (Q_{\pi}, Q_{\pi'})} \frac{1}{P_{\pi,\pi'}(p^{-s})},\] where $n,n'$ are the degrees of $\pi, \pi'$ respectively and $Q_{\pi}, Q_{\pi'}$ are their conductors, and $P_{\pi,\pi'}(p^{-s})$ are polynomials in $p^{-s}$ of degree at most $nn'$ that are complicated to describe but depend sensitively on $\pi$ and $\pi'$.
In particular cases, this can be written in a simpler form. If $\pi,\pi'$ are modular forms (so of degree $2$) of squarefree level $q_{\pi}, q_{\pi'}$ and character $\chi_{\pi}, \chi_{\pi'}$, then \[L(s,\pi \otimes \pi') = L(2s, \chi_{\pi} \chi_{\pi'}) \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}.\] Note the additional term $L(2s, \chi_{\pi} \chi_{\pi'})$ out the front; this is because the Dirichlet series $\sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}$ isn't an $L$-function (it doesn't satisfy a functional equation of the type you'd want). In general (i.e. for higher $n,n'$), the part of $L(s,\pi \otimes \pi')$ that $\sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}$ is "missing" is a little more complicated.
Similarly, the archimedean gamma factors $\gamma(s, \pi \otimes \pi')$ can be a little more complicated than the naïve guess. Nevertheless, here's an example that answers your question:
Take $\pi = f$ and $\pi' = g$ to be Maass newforms both of weight $0$ and both even or both odd, with Laplacian eigenvalue $1/4 + t_f^2$ and $1/4 + t_g^2$. Then \[\gamma(s,f) = \pi^{-s} \Gamma\left(\frac{s + \kappa + it_f}{2}\right) \Gamma\left(\frac{s + \kappa - it_f}{2}\right), \qquad \gamma(s,g) = \pi^{-s} \Gamma\left(\frac{s + \kappa + it_g}{2}\right) \Gamma\left(\frac{s + \kappa - it_g}{2}\right),\] where $\kappa = 0$ if $f$ and $g$ are both even and $1$ if they're both odd. We also have that \[\gamma(s,f \otimes g) = \pi^{-2s} \Gamma\left(\frac{s + i(t_f + t_g)}{2}\right) \Gamma\left(\frac{s + i(t_f - t_g)}{2}\right) \Gamma\left(\frac{s - i(t_f + t_g)}{2}\right) \Gamma\left(\frac{s - i(t_f - t_g)}{2}\right).\] Finally, we note that there do exist Maass newforms of weight $0$ that are even and have Laplacian eigenvalue $1/4$, so that $t_f = t_g = 0$. This gives an affirmative answer to your question.