Case 3:
Let me define $t=x\sqrt{2\gamma}$, then it is known from random-matrix theory (see, for example, Forrester's book) that for a fixed $\gamma$ the probability distribution $P(x_1)$ of a single eigenvalue $x_1$ tends in the limit $n\rightarrow\infty$ to the $\gamma$-independent semicircle $$P(x)=\frac{1}{\pi n}\sqrt{2n-x^2},\;\;|x|\leq\sqrt{2n}.$$ The desired ratio $\nu$ then evaluates to $$\nu=\frac{\int (2\gamma x^2-1)P(x)\,dx}{\left[\int (2\gamma x^2-1)^2P(x)\,dx\right]^{1/2}}=\frac{\gamma n-1}{\sqrt{2 \gamma n (\gamma n-1)+1}}\rightarrow \frac{1}{\sqrt 2}\;\;\text{for}\;\;n\rightarrow\infty.$$
Case 2:
The case that $n\rightarrow\infty$, $\gamma\rightarrow 0$ at fixed $\gamma n=\alpha>0$ has been studied in The mean spectral measures of random Jacobi matrices related to Gaussian beta ensembles (2014), see also arXiv:1611.09476. The probability distribution $P_\alpha(\lambda)$ is given in this limit by $$P_\alpha(\lambda)=\frac{e^{-\lambda^2/2}}{\sqrt{2\pi}}\frac{1} {|f(\lambda)|^2},\;\;f(\lambda)=\sqrt{\frac{\alpha}{\Gamma(\alpha)}}\int_0^\infty t^{\alpha-1}e^{i\lambda t-t^2/2}\,dt$$ $$\Rightarrow P_\alpha(\lambda)=\frac{2^{\frac{3}{2}-{\alpha}} e^{-\frac{{\lambda}^2}{2}} \Gamma ({\alpha})}{ \sqrt{\pi } {\alpha}\Gamma \left(\frac{{\alpha}}{2}\right)^2 \, _1F_1(\frac{{\alpha}}{2};\frac{1}{2};-\frac{{\lambda}^2}{2}){}^2+2\sqrt{\pi } {\alpha} {\lambda}^2 \Gamma \left(\frac{{\alpha}+1}{2}\right)^2 \, _1F_1(\frac{{\alpha}+1}{2};\frac{3}{2};-\frac{{\lambda}^2}{2}){}^2}.$$ For $\alpha=1$ this simplifies to $$P_1(\lambda)=\frac{\sqrt{2}\pi^{-3/2} e^{\lambda^2/2}}{\text{erfi}\left(\lambda/\sqrt{2}\right)^2+1}.$$ From the desired $\nu$ can be readily computed (I will do so shortly).