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
$$\sup_{t \leq T} \mathbb{E}(X_t^3) < \infty, \qquad T>0.$$
You have some (typo ?) issue in your definition. An Itô process $(I_t)$ is a stochastic process which can be written as $$X_t = X_0 + \int_{s=0}^t \mu_s ds + \int_{s=0}^t\sigma_sdB_s $$
Where $(B_t)$ is a standard Brownian motion and $(\mu_t)$ (drift) and $(\sigma_t)$ (diffusion) are two stochastic processes adapted to $(B_t)$.
As two extremely elementary examples of Itô processes, you can think of the Brownian motion $(B_t)$, which can be written as $B_t =0 + \int_{s=0}^t 0\ ds + \int_{s=0}^t1\ dB_s$, or the deterministic process $(t)$, which writes $ t = 0 + \int_{s=0}^t 1\ ds + \int_{s=0}^01\ dB_s $.
Now Itô's lemma states that, given an Itô process $(X_t)$ with drift $ (\mu_t)$ and diffusion $ (\sigma_t)$, and a twice differentiable function $f(\cdot,\cdot)$, the process $(f(t,X_t)) $ is also an Itô process with drift $ (\mu_t^f) = \left({\frac {\partial f}{\partial t}}(t,X_t)+\mu _{t}{\frac {\partial f}{\partial x}}(t,X_t)+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}(t,X_t)\right)$ and diffusion $(\sigma_t^f) = \sigma _{t}{\frac {\partial f}{\partial x}}(t,X_t)$.
In other words,
$$f(t,X_t) = f(X_0) + \int_{s=0}^t \mu_s^f ds + \int_{s=0}^t\sigma_s^f\ dB_s \tag1$$
To apply that in the setting of example a), we identify :
$(B_t)$ is an Itô process with drift $0$ and diffusion $1$, $(X_t) = (B_t^2) = f(t,B_t)$ with $f(t,x) := x^2$. As you did, we compute every derivative term that appear in the formula of Itô's lemma :
$$\frac{\partial f}{\partial t}(t,B_t) =0,\; \frac{\partial f}{\partial x}(t,B_t) = 2B_t,\; \frac{\partial^2 f}{\partial x^2}(t,B_t) =2$$
Now we can plug the values in to get $\mu^f$ and $\sigma^f$ :
$$\mu^f_t = 0 + 0 + 1 = 1 \;\text{ and } \sigma^f_t = 1\times2B_t = 2B_t $$
We can finally plug in these expressions in $(1)$ to get the solution :
$$\begin{align}B_t^2 = f(t,B_t) &= B_0^2 + \int_{s=0}^t \mu_s^f ds + \int_{s=0}^t\sigma_s^f\ dB_s\\
&=0+ \int_{s=0}^t 1\ ds + \int_{s=0}^t2B_s dB_s\\
&=\int_{s=0}^t \ ds + \int_{s=0}^t2B_s dB_s \ \; \; \; \; \blacksquare \end{align} $$
If you understood this, you can proceed the same exact way to solve example b). I let you do the calculations yourself ;)
Best Answer
When working with these computations, it's usually easier to write the SDE in differential form. We know that the dynamics of $X_t$ are given by $$X_t = f(t) dB_t - \frac{1}{2}f^2(t) dt$$
Applying Itô's lemma with $g(x) = e^x$, we get: $$\begin{align*} dY_t &= \frac{1}{2} e^{X_t}(dX_t)^2 + e^{X_t} dX_t \\ &= \frac{1}{2}e^{X_t} f^2(t)dt + e^{X_t}f(t)dB_t - \frac{1}{2} e^{X_t} f^2(t)dt \\ &= e^{X_t}f(t) dB_t \end{align*}$$
Since $Y_t$ is driftless, it is a local martingale. But then, if $f$ is reasonable, we actually have that $E\left( \int_0^t e^{2X_s} f^2(s)ds \right) < \infty$, which lets you conclude that $Y_t$ is a martingale.