The following horrible formula for the joint distribution of max, min and end value of a Brownian motion was copied without guarantees from the Handbook Of Brownian Motion (Borodin/Salminen), 1.15.8, p.271. First, for simplicity, this is only written for $\sigma=1,t=1$, and the more general case comes directly from scaling. If we shorten W as the Brownian Botion at t=1, m as the minimum and M as the maximum over $[0,1]$, then for $a < min(0,z) \le max(0,z) < b$ it holds
$$
P(a < m, M < b, W \in dz) = \frac{1}{\sqrt{2\pi}}e^{(\mu z-\mu^2/2)} \cdot \sum_{k =-\infty}^{\infty} \Bigl(e^{-(z+2k(b-a))^2/2} - e^{(z-2a + 2k(b-a))^2/2} \Bigr) dz\; .
$$
(Apologies for using z here in a different context.) If one really wants to, one can compute from this an even more horrible formula for the above probability. It is now in principle possible to derive from this a formula for what you want, by finding the density function $p_{m,M,W}$, and using
$$
P(e^M-e^m\le r) = \int_{(x,y,z)\ :\ e^x \le e^z \le e^y \le e^x + r} p_{m,M,W}(x,y,z) d(x,y,z)\;,
$$
but I shudder at the monster I expect to fall out from this. It might be better to give up and simulate the probability in question, and find some asymptotics.
However, if you would like to proceed with it, I suggest you look not into the Handbook Of Brownian Motion, but rather into this paper, as it is much more readable.
Yes, the set $\{\omega\mid GBM(t)(\omega)\to 0\}$ has probability 1.
This is not a contradiction to "there is no drift", because at every finite time you have a (small) nonzero probability of the stock price being sky-rocketed to compensate for much larger probability that the stock is almost worthless. Note that "there is no drift" means what happens to expectation, but you can't push the limit inside the integral because the hypotheses of dominated convergence theorem fails.
A discrete-time martingale should help you understand this. Suppose the stock price either become worthless or double on each time step, both with probability $\frac12$. So it is a martingale (therefore has no drift), but the price is almost surely 0 in the long run.
Best Answer
You are on the right track so far, but you have the drift wrong. The drift of $\log(X(t))$ is, from Ito's lemma, $-\frac{1}{2}\sigma^2.$
After that you use the fact that since it's a brownian motion starting from $\log(8.0)$ the distribution is $$\log(X(t)) \sim N\left(\log(8)-\frac{1}{2}\sigma^2t,\sigma^2t\right).$$ So let $Z$ be normal with mean $\log(8)-\frac{1}{2}\sigma^2\left(\frac{1}{2}\right)$ and variance $\sigma^2\left(\frac{1}{2}\right).$ To finish the problem, you need to compute $$ P(Z>\log(8.40)).$$