The precise, mathematical statement of the uncertainty principle is $\sigma^2_x \sigma^2_k \geq 1/4$. The use of deltas is just an informal way of talking about it. Nevertheless, it's pretty common to say, for instance, that the width of a peak is either the standard deviation or some quantity proportional to it--see, for example, full width at half maximum, which ends up being about $2.35\sigma$.
I'm not really sure what a slit would look like in 1 dimension. It's easier for me to consider a particle in a 1d infinite square well. Note that the infinite well absolutely forbids any leakage of the particle into the forbidden region, just like the classical case. In this case, the variances depend on the energy of the particle. For a particle in one of the $n$th energy eigenstate of an infinite well with width $L$, the variances are (per wikipedia)
$$\sigma^2_x = \frac{L^2}{12} \Bigg ( 1 - \frac{6}{n^2 \pi^2} \Bigg), \quad \sigma_k^2 = \frac{n^2 \pi^2}{L^2}$$
The product of the variances is then $\sigma_x^2 \sigma_k^2 = (n^2 \pi^2/12 - 1/2)$. For $n=1$, this is about $.322 \geq .25$, as required.
You can't really see what the uncertainties will be by inspection, by the geometry of the problem. These are the uncertainties for energy eigenstates, and there's no reason to expect that a particle will be in an eigenstate (which would then make the computation more complicated).
Really, one just calculates the variances of the wavefunction with respect to $x$ and $k$. You might be able to get a rough idea from the quantities in the problem (for instance, the standard deviation with respect to $x$ is indeed proportional to $L$, but only proportional, not exactly $L$), but that's all.
You ask if a particle has nonzero probability of existing everywhere. To be pedantic, a particle has zero probability of existing at any specific point, but it typically has a nonzero probability of existing in a region of any finite size. This infinite square well is an exception, as the infinite potential around the box absolutely forbids particles.
Uncertainty really is just a loose, loose word to use. It almost always really means standard deviation of the wavefunction.
Heisenberg's uncertainty principle is
$$\Delta x \Delta p \geq \hbar/2.$$
Since the well is of width $L$, you have a measure for the uncertainty on the position $\Delta x$. Then assume the lowest possible value for $\Delta p$, i.e. the one for which the above inequality becomes an equality. Lastly, use $E = \dfrac{p^2}{2m}$ to find an expression for $E$.
A useful question to look at as well might be this one.
Best Answer
One way to think of this is in terms of expectation values. When you say $\Delta p$, what you really mean is the standard deviation of $p$.
$$ \Delta p = \sqrt{\langle p^2\rangle-\langle p \rangle^2} $$ In the case of the ground state, you expect $\langle p \rangle=0$ by symmetry, so you just have $\Delta p = \sqrt{\langle p^2\rangle}$. Then you can consider the expectation value of the energy,
$$ \langle E\rangle = \langle \frac{p^2}{2m}\rangle=\frac{\langle p^2\rangle}{2m}=\frac{(\Delta p)^2}{2m} $$
So far, everything we've written has been exact. But we want to find the minimum possible value for the energy. A moments thought should tell you that the minimum of $\langle E\rangle$ and the minimum of the energy coincide. So you try to find the smallest possible $\langle E\rangle$, and call that $E_{\min}$. That means you want to find the smallest possible $\Delta p$. But of course you know $\Delta x \lesssim a$, so the smallest $\Delta p$ is $\tilde{}\frac{h}{2\pi a}$. Plugging that in gives you $E_\min$.
They key is realizing that if $\langle p\rangle=0$, then the expectation value of $p^2$ is exactly $(\Delta p)^2$. Of course, everything after that is just approximations, but sometimes they work pretty well!