If I want to find y such that P(Y< y)=0.999 of Gumbel Distribution with a=30.134711, b= 7.621868. where $$f(y\,;a,b)=\frac{1}{b}\exp\left[-\frac{(y-a)}{b}-e^{-(y-a)/b}\right]\quad,\,y\in\mathbb R\quad,\,a\in\mathbb R,b>0$$
Is it correct that I use quantile of Gumbel distribution with p=0.999 to find such y which is =82.7809
Yes, that looks correct
From that density, you have $$\mathbb P(Y \le y)=F(y)= \exp(-\exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-b\log_e(-\log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$