How to work with random variables? 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 $t$.
 A: 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.
