I will first make a few simplifications. Define $G_t := F_t - \alpha$. Then of course $dG_t = dF_t$. So then we are solving for $dG_t = \beta_tG_tdW_t$. Since $t\mapsto\beta_t$ is deterministic, this is still geometric Brownian motion. If $t\mapsto\beta_t$ is Lipschitz, which you would assume to guarantee uniqueness, then $G_t$ has the Markov property. So then, we can look at the solution
$$G_t = G_0e^{-\frac{1}{2}\int_0^t\beta_s^2\,ds + \int_0^t\beta_s\,dW_s}$$
instead of arbitrary $0 \leq t\leq T$.
Define the function $f_t = \int_0^t\beta_s^2\,ds$ and the process $X_t = \int_0^t\beta_s\,dW_s$. Hence, $G_t = G_0e^{-\frac{1}{2}f_t + X_t}$. Note that $dX_t = \beta_tdW_t$ and $d[X,X]_t = \beta^2_tdt$. Since you wanted a full analytical solution, I will be a bit pedantic and treat $G_0$ as another stochastic process (constant over time but still random), i.e. $Y_t = G_0$ for all $t \geq 0$. Then of course $dY_t = 0$ and $d[Y,Y]_t = 0$. I hope it is also clear to you that $d[X,Y]_t = 0$ since all the paths of $Y$ are constant.
Now consider the function $r : (t,x,y) \mapsto ye^{-\frac{1}{2}f_t + x}$. You can see that $G_t = r(t,X_t,Y_t)$. This function is infinitely differentiable in all its arguments so we have more than what we need to apply Ito's lemma. I will use subscripts to denote partial derivatives of $r$ below.
$$dG_t = r_t(t,X_t,Y_t)dt + r_x(t,X_t,Y_t)dX_t + r_y(t,X_t,Y_t)dY_t + \frac{1}{2}r_{xx}(t,X_t,Y_t)d[X,X]_t + \frac{1}{2}r_{yy}(t,X_t,Y_t)d[Y,Y]_t + r_{xy}(t,X_t,Y_t)d[X,Y]_t$$
The last two terms and the $dY$ term drop due to the considerations above.
$r_t = -\frac{1}{2}\frac{df}{dt}r = -\frac{1}{2}\beta^2_tr$, $r_x = r$ and $r_{xx} = r$. Putting all these together,
$$dG_t = -\frac{1}{2}\beta^2_tG_tdt + G_t \beta_tdW_t + \frac{1}{2}G_t\beta^2_tdt = G_t \beta_tdW_t$$
So then the given solution satisfies the SDE. For the other question we need to look at
$$P\{F_T \leq x\} = P\{\alpha + (F_0-\alpha)e^{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u} \leq x\} $$
$$P\{F_T \leq x\} = P\{e^{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u} \leq \frac{x-\alpha}{F_0 -\alpha}\}$$
$$P\{F_T \leq x\} = P\{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u \leq \log\left(\frac{x-\alpha}{F_0 -\alpha}\right)\}$$
$$P\{F_T \leq x\} = P\{\int_0^t\beta_u\,dW_u \leq \log\left(\frac{x-\alpha}{F_0 -\alpha}\right) + \frac{1}{2}\int_0^t\beta_u^2\,du\}$$
Note that $\int_0^t\beta_u\,dW_u \sim \mathsf{N}(0,\int_0^t\beta_u^2\,du)$. Then we can write
$$P\{F_T \leq x\} = \Phi\left(\frac{\log\left(\frac{x-\alpha}{F_0 -\alpha}\right) + \frac{1}{2}\int_0^t\beta_u^2\,du}{\sqrt{\int_0^t\beta_u^2\,du}}\right)$$
Here $\Phi$ is the distribution function of a standard normal random variable. The sensitivity of this expression with respect to $\alpha$ you can judge for yourself by looking at the derivatives or using numerical methods.
One last remark: In the second part I conveniently assumed that $F_0$ is a constant. $F_0$ can be a random variable itself. Note that for $F_0$ to be measurable with respect to $\mathcal{F}_0$ it would have to be independent of $W$. Then, all you need to specify is the marginal distribution of $F_0$ to evaluate the probability $P\{F_T \leq x\}$ (rather than the joint distribution of $\int_0^t\beta_u\,dW$ and $F_0$).
