Instead of using exp, I'll show how to derive $\int_0^t f(B_t)\,dB_t$, from the definition, assuming $\,f''$ exists and is continuous.
Let $F(x)=\int_0^xf(s)\,ds$. We of course start with a partition $0=t_0<t_1<\dots<t_m=t$, and setting $B_k=B_{t_k}$. Let $\mu=\sup_k t_{k+1}-t_k$ be the mesh of the partition.
Write
$$
F(B_{k+1})=F(B_k)+f(B_k)(B_{k+1}-B_k)+\frac12f'(B_k)(B_{k+1}-B_k)^2+R_k(B_{k+1})
$$
where $R_k(x)$ is the remainder term for the Taylor series centered at $B_k$. Rearranging, and letting $\Delta B_k=B_{k+1}-B_k$,
$$
f(B_k)\Delta B_k=F(B_{k+1})-F(B_{k})-\frac12 f'(B_k)(\Delta B_k)^2-R_k(B_{k+1})
$$
We then sum the above expression over $0\le k\le m-1$, and take the limit in probability as $\mu\to 0$ (meaning we are really using a sequence of partitions). The LHS of above becomes $\int f(B_t)dB_t$, by definition.
On the right, the $F(B_{k+1})-F(B_{k})$ part is telescoping, adding up to $F(t_n)-F(t_0)=F(t)$.
To evaluate the $f'(B_k)(\Delta B_k)^2$ part, we let $\Delta t_k=t_{k+1}-t_k$, and replace this with
$$
f'(B_k)\Delta t_k + f'(B_k)((\Delta B_k)^2 - \Delta t_k)
$$
Then $\sum_{k=0}^{m-1}f'(B_k)\Delta t_k \to \int_0^t f'(B_k)dt$, and $\sum_{k=0}^{m-1}f'(B_k)(\Delta B_k - \Delta t_k)\to 0$. You show the last limit approaches zero by computing its variance:
$$
Var\left(\sum_{k=0}^{m-1}f'(B_k)((\Delta B_k)^2 - \Delta t_k)\right)=\sum_{k=0}^{m-1}Var(f'(B_{k}))(\Delta t_k)^2 Var\left(\frac{\Delta (B_k)^2}{\Delta t_k}-1\right)$$
$$\qquad \qquad \qquad\le Ct\cdot \mu\cdot \kappa\cdot \sum \Delta t_k=Ct\mu \kappa t\stackrel{\mu\to 0}\to 0$$
where $C=\sup_{s\in [0,t]} f'(s)$, so that $Var(f'(B_k))\le Var(CB_k)\le C^2 t_k\le C^2t$, and $\kappa$ is the variance of a $\chi^2$ random variable (note that $\frac{\Delta (B_k)^2}{\Delta t_k}$ are i.i.d. $\chi^2$). Since the variance $\to0$, and its mean is zero, the LHS $\to 0$ in $L_2$.
Finally, letting $K=\sup_{0\le s\le t}f''(t)$, the remainder term $R(B_{k+1})$ is uniformly bounded by $K(\Delta B_k)^3$, and a similar variance calculation shows that $\sum_{k=0}^{m-1} R_k(B_{k+1})\to 0$.
Putting this all together,
$$
\int_0^t f(B_k)\,d B_t=F(B_t)-\frac12\int_0^t f'(B_k)\,dt
$$
If $X_t $ is of the form
$$
X_t = \int_0^t f(s,\omega)dB_s(\omega),\quad t\geq 0,
$$ then by using localization technique, we may assume $X_t$ is a $L^2$-bounded martingale. In that case, there is a sequence of elementary processes $\{f_n\}$ such that
$$
E[\int_0^T |f_n(s,\cdot)-f(s,\cdot)|^2ds]\to 0
$$ for all $T>0$. Let $X_{n,t} = \int_0^t f_n(s,\omega)dB_s(\omega)$. Then by martingale maximal inequality, we have
$$
E[\sup_{t\in [0,T]}|X_t -X_{n,t}|^2]\leq C E[\int_0^T |f_n(s,\cdot)-f(s,\cdot)|^2ds]\to 0.
$$ This implies that $X_{n.t} \to_p X_t$ uniformly on every compact interval $[0,T]$. We may further assume that $g\in C^2_0$ via localization method. Then,
$$
g(X_{n,t}), g'(X_{n,t}), g''(X_{n,t})
$$ converge in probability to
$$
g(X_t), g'(X_t), g''(X_t)
$$ locally uniformly.
Best Answer
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 ;)