If $X$ and $Y$ are independent random variables described by standard normal distribution could you please explain how to formally evaluate probabilities of occurrences such as $X-Y>0$ (intuitively it's $0.5$ of course) or $X^2-Y^2>0$?
Ultimately I'd like to be able to tell how likely is it that parabola $z(t)=X^2+2Yt+t^2$ has a root in positive semi-axis of $t$.
If $X$ and $Y$ are jointly normal (and not necessarily independent), then $X-Y$ is a normal random variable. More generally, $aX+bY$ is a normal random variable with mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y$ where $\rho$ is the correlation coefficient. For your special case, $X-Y \sim \mathcal N(0,2)$ and so, with $\Phi(\cdot)$ denoting the cumulative probability density function of the standard normal random variable, $$P\{X-Y > 0\} = 1-\Phi\left(\frac{0}{\sqrt{2}}\right) = 1 - \frac{1}{2} = \frac{1}{2}$$ just as you intuited, presumably via the symmetry argument that since $X$ and $Y$ are independent identically distributed random variables, then $P\{X<Y\}=P\{X>Y\}$. But the formal approach will work in more general cases too. For example, more generally, $$P\{aX+bY \leq c\} = \Phi\left(\frac{c-(a\mu_X+b\mu_Y)}{\sqrt{a^2\sigma_X^2+b^2\sigma_Y^2+2ab\rho\sigma_X\sigma_Y}}\right).$$ Finding $P\{X^2-Y^2 < 0\}$ requires more computation in the general case, but for the case when $X$ and $Y$ are independent identically distributed normal random variables, $X^2$ and $Y^2$ also are independent random variables with identical (though non-normal) distribution, and so $$\{X^2-Y^2 < 0\} = \frac{1}{2}$$ by symmetry. Note that it is not necessary to find the distributions of $X^2$ and $Y^2$ and nor is integration of the joint density over a region necessary. An alternative calculation is to note that $\{X^2 - Y^2 < 0\}$ if and only if $X+Y$ and $X-Y$ have opposite signs, and since $X+Y$ and $X-Y$ happen to be independent $\mathcal N(0,2)$ random variables, the probability is $\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2} = \frac{1}{2}$.
With regard to your third question, the parabola $z = X^2+2Yt+t^2$ has value $X^2 \geq 0$ at $t=0$. If the slope at $t=0$, $\left.\frac{dz}{dt}\right|_{t=0} = 2Y+2t\bigr|_{t=0} = 2Y $ is also positive, then the parabola will not cross the positive $t$ axis. So, $Y < 0$ is a necessary condition for getting a crossing on the positive $t$ axis. Also, $z$ has a minimum value of $X^2-Y^2$ at $t = -Y$. Thus, the parabola crosses the positive $t$ axis (twice) exactly when $\{Y < 0\}$ and $\{X^2-Y^2 < 0\}$. The probability is thus $\frac{1}{4}$ via the circular symmetry of the joint density of two independent $\mathcal N(0,1)$ random variables.
