It's not difficult to see that
$$X_t := \exp \left(\sqrt{2} B_t \right)$$
solves the given SDE. (You can either use Itô's formula to check it or use some standard methods for linear SDE's to obtain this solution.) Moreover, by Ito's formula:
$$f(t,X_t)-\underbrace{f(0,x_0)}_{x_0} = \sqrt{2} \int_0^t e^{-s} \cdot X_s \, dB_s + \int_0^t e^{-s} \cdot X_s + (-e^{-s} \cdot X_s) \, ds \\ = \sqrt{2} \int_0^t e^{-s} \cdot e^{\sqrt{2} B_s} \,dB_s \\ \Rightarrow f(t,X_t) = \underbrace{\sqrt{2} \int_0^t e^{\sqrt{2} B_s-s} \,dB_s}_{M_t} + \underbrace{x_0}_{A_t}$$
where $x_0=1$. Let
$$g(s,w) := \sqrt{2} \cdot e^{\sqrt{2} B(s,w)-s}$$
Then $g \in L^2(\lambda_T \otimes \mathbb{P})$, i.e.
$$\int_0^T \int_\Omega g(s,w)^2 \, d\mathbb{P} \, ds <\infty$$
There is a general result which says that this condition implies that $M_t$ is a martingale (and not only a local one). Moreover,
$$\langle M,M \rangle_t = \int_0^t |g(s,w)|^2 \, ds$$
(see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 14.13).
Concerning the integral $\mathbb{E}(e^{-\tau} \cdot X_\tau)$: Remark that
$$\tau = \inf\{t \geq 0; X_t=2-t\} = \inf\{t \geq 0; \sqrt{2} B_t = \ln(2-t)\}$$
Now let
$$\sigma := 2\tau = \inf\{t \geq 0; \underbrace{\sqrt{2} B_{\frac{t}{2}}}_{=:W_t} = \ln (2-t/2)\}$$
where $(W_t)_{t \geq 0}$ is again a Brownian Motion (scaling property). Thus
$$\mathbb{E}(e^{-\tau} \cdot X_\tau) = \mathbb{E}(e^{-\tau+\sqrt{2} B_\tau}) = \mathbb{E}(e^{-\frac{\sigma}{2}+W_\sigma}) \stackrel{\ast}{=} 1$$
In $(\ast)$ we applied the exponential Wald identity (see remark).
Remark Exponential Wald identity: Let $(W_t)_{t \geq 0}$ a Brownian motion and $\sigma$ a $\mathcal{F}_t^W$-stopping time such that $\mathbb{E}e^{\sigma/2}<\infty$, then $\mathbb{E}(e^{W_\sigma-\frac{\sigma}{2}})=1$. (see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 5.14)
Yes, the process $(X_t)$ defined by $X_t=\displaystyle\int_0^tB_s^2\mathrm dB_s$ is a martingale, as every square integrable stochastic integral $\displaystyle\int_0^tu(s,B_s)\mathrm dB_s$, but, no, going back to Itô's formula to show this is not useful since this is going backwards because one already knows that $\mathrm dX=B^2\mathrm dB$, which is the kind of conclusion Itô's formula can help to reach.
No, the process $(Y_t)$ defined by $Y_t=B_t^3$ is not a martingale since, for every $t\gt s\gt0$, the decomposition $B_t=B_s+(B_t-B_s)$ with $B_s$ measurable with respect to $\mathcal F_s$ and $B_t-B_s$ independent of $\mathcal F_s$ yields the almost sure identity $E(B_t^3\mid\mathcal F^B_s)=B_s^3+3(t-s)B_s\ne B_s^3$.
No, a process being a supermartingale does not imply that it is a martingale, and no, the process $-\frac13B^3$ is not a supermartingale (and not a submartingale either).
Best Answer
There are general statements on the existence of moments of solutions to SDEs. If you are lucky, you have one of those results at your disposal to deduce that $\sup_{s \leq t} \mathbb{E}(|X_s|^3)<\infty$ for all $t>0$. If this is not the case, then you need to do some further calculations.
In the following, I will consider $\mathbb{E}(X_t^4)$ (rather than $\mathbb{E}(|X_t|^3)$) since this saves us the trouble to deal with the modulus. By Itô's formula, we have
$$X_t^4 = 8 \int_0^t X_s^3 \sqrt{X_s} \, dB_s +30 \int_0^t X_s^3 \, ds.$$
Now we want to take the expectation, but since we do not yet know whether the expectation is finite, we first need to some stopping. To this end, define
$$\tau_r := \inf\{t \geq 0; |X_t| \geq r\},$$
note that $\tau_r \uparrow \infty$ as $r \to \infty$ and $|X_{t \wedge \tau_r}| \leq r$. Since
$$t \mapsto \int_0^{t \wedge \tau_r} X_s \sqrt{X_s} \, dB_s$$
is a martingale (because of the stopping!), and, hence, has expectatio zero, we get
$$\mathbb{E}(X_{t \wedge \tau_r}^4) = 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} X_s^3 \, ds \right).$$
Using the elementary estimate
$$x^3 \leq |x|^3 \leq 1+x^4, \qquad x \in \mathbb{R},$$
for $x=X_s$, we find that
$$\mathbb{E}(X_{t \wedge \tau_r}^4) \leq 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} (1+X_s^4) \, ds \right) \leq 30 \int_0^t \mathbb{E}(1+X_{s \wedge \tau_r}^4) \, ds.$$
If we define $u(t) := \mathbb{E}(1+X_{t \wedge \tau_r}^4)$ for fixed $r>0$, then
$$u(t) \leq 1+ 30 \int_0^t u(s) \, ds$$
and so Gronwall's lemma yields
$$u(t) \leq c e^{Mt}$$
for suitable constants $c,M>0$, which do not depend on $r>0$. Consequently, we have shown that
$$\mathbb{E}(1+X_{t \wedge \tau_r}^4) \leq c e^{Mt}, \qquad t \geq 0.$$
By Fatou's lemma, this implies
$$\mathbb{E}(1+X_{t}^4) \leq c e^{Mt}, \qquad t \geq 0.$$
In particular,
$$\sup_{t \leq T} \mathbb{E}(X_t^4) < \infty, \qquad T>0,$$
which also yields
$$\sup_{t \leq T} \mathbb{E}(X_t^3) < \infty, \qquad T>0.$$