First of all, note that the natural filtration of $(X_t)_{t \geq 0}$ equals the natural filtration of $(W_t)_{t \geq 0}$; this follows directly from the relation
$$X_t = \exp \left( \left[ \mu - \frac{\sigma^2}{2} \right] t + \sigma W_t \right).$$
The Markov property of $(X_t)_{t \geq 0}$ can be proved as follows: Fix some bounded Borel-measurable function $f$ and $s \leq t$. For brevity, set $c := \mu - \frac{\sigma^2}{2}$. Then
$$\begin{align*} \mathbb{E}(f(X_t) \mid \mathcal{F}_s) &= \mathbb{E} \bigg( f \left[ e^{ct} e^{\sigma(W_t-W_s)} e^{\sigma W_s} \right] \mid \mathcal{F}_s \bigg). \end{align*}$$
Since $W_t-W_s$ is independent from $\mathcal{F}_s = \sigma(W_r; r \leq s)$ and $W_s$ is $\mathcal{F}_s$-measurable, we get
$$\mathbb{E}(f(X_t) \mid \mathcal{F}_s) = g(e^{ct} e^{\sigma W_s}) \tag{1}$$
where
$$g(y) := \mathbb{E} \left( f(y e^{\sigma (W_t-W_s)}) \right).$$
Since the right-hand side of $(1)$ is $X_s$-measurable, the tower property yields
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E} \bigg[ \mathbb{E}(f(X_t) \mid \mathcal{F}_s) \mid X_s \bigg] \stackrel{(1)}{=} g(e^{ct} e^{\sigma W_s}). \tag{2}$$
Combining $(1)$ and $(2)$ we find
$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E}(f(X_t) \mid \mathcal{F}_s).$$
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 $$ P(Z>\log(8.40)).$$
Best Answer
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.