Note that $$\begin{align*}B_T^3 &=\sum_{j=1}^n B_{s_j}^3 - B_{s_{j-1}}^3 = \sum_{j=1}^n (B_{s_{j-1}}+(B_{s_j}-B_{s_{j-1}}))^3-B_{s_{j-1}}^3 \notag \\ &= \underbrace{\sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3}_{=: I_1}+ 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{=:I_2} + 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}})}_{=: I_3} \tag{1}\end{align*}$$
We consider the terms separately and start with $I_3$:
For \begin{equation*} g^\Pi(s,w) := \sum_{j=1}^n \underbrace{B_{s_{j-1}}^2}_{\in L^2} \cdot 1_{[s_{j-1},s_j)}(s) \end{equation*}
we have \begin{align*} \mathbb{E} \left( \int_0^T |g^\Pi(s)-B_s^2|^2 \, ds \right) &\stackrel{\text{Fub}}{=} \sum_{j=1}^n \int_{s_{j-1}}^{s_j} \mathbb{E}(|B_{s_{j-1}}^2-B_s^2|^2) \, ds \\ &\leq \sup_{|s-t| \leq |\Pi|} |B_s^2-B_t^2|^2 \sum_{j=1}^n \int_{s_{j-1}}^{s_j} 1 \ \, ds \\ &\to 0 \qquad (|\Pi| \to 0)\end{align*}
where $|\Pi| := \max (s_j-s_{j-1})$ denotes the mesh size of the partition $\Pi = \{0=s_0<\ldots<s_n=T\}$. In the last step, we have used that $[0,T] \ni t \mapsto B_t$ is uniformly continuous (almost surely). From the definition of the Itô integral, it follows that
$$g^\Pi \bullet B_T = \sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}}) \stackrel{L^2}{\to} \int_0^T B_s^2 \, dB_s $$
Hence, $$I_3 \stackrel{|\Pi| \to 0}{\to} \int_0^T B_s^2 \, dB_s. \tag{2}$$
Next we consider $I_2$. To this end, let $X_j := B_{s_{j-1}} \cdot ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1})$. Then $$\begin{align*} \left\| \left( \sum_{j=1}^n X_j \right)^2 \right\|_{L^2(\mathbb{P})}^2 &\stackrel{\ast}{=} \sum_{j=1}^n \mathbb{E}(X_j^2) = \sum_{j=1}^n \mathbb{E}((B_{s_{j-1}}^2 \cdot ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1})))^2) \\ &= \sum_{j=1}^n \underbrace{\mathbb{E}(B_{s_{j-1}}^2)}_{s_{j-1}} \cdot \underbrace{\mathbb{E}( ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1}))^2)}_{\stackrel{L1.6}{=} (s_j-s_{j-1})^2} \\ &= \sum_{j=1}^n s_{j-1} \cdot (s_j-s_{j-1})^2 \leq |\Pi| \cdot \underbrace{\sum_{j=1}^n s_{j-1} \cdot (s_j-s_{j-1})}_{\leq T^2} \\ &\to 0 \qquad (|\Pi| \to 0) \end{align*}$$
For $(\ast)$ note that \begin{align*} \mathbb{E}(X_j \cdot X_k) &= \mathbb{E}(\mathbb{E}(X_j \cdot X_k|\mathcal{F}_{k-1})) \\ &= \mathbb{E}(X_j \cdot B_{s_{k-1}} \cdot \underbrace{\mathbb{E}((B_{s_j}-B_{s_{j-1}})^2)-(s_j-s_{j-1})|\mathcal{F}_{k-1})}_{= \mathbb{E}((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1}) = 0}) = 0 \end{align*} for all $j>k$ where we have used that $B_{s_j}-B_{s_{j-1}}$ is independent from $\mathcal{F}_{k-1}:=\mathcal{F}_{s_{k-1}}$. Consequently, there exists a subsequence converging almost surely to $0$. Since
$$\sum_{j=1}^n B_{s_{j-1}} \cdot (s_j-s_{j-1}) \to \int_0^T B_s \, ds \quad \text{a.s.} \quad (|\Pi| \to 0)$$ we conclude $$I_2 \to \int_0^T B_s \, ds \quad \text{a.s.} \quad (|\Pi| \to 0) \tag{3}$$
It remains to determine $\lim_{|\Pi| \to 0} I_1.$ To this end, we note that $$\begin{align*} \mathbb{E} \left( \left| \sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3 \right| \right) &\leq \sum_{j=1}^n \mathbb{E}(|B_{s_j}-B_{s_{j-1}}| \cdot |(B_{s_j}-B_{s_{j-1}})^2|) \\ &\stackrel{\text{CSI}}{\leq} \sum_{j=1}^n \underbrace{\|B_{s_j}-B_{s_{j-1}}\|_{L^2(\mathbb{P})}}_{\sqrt{s_j-s_{j-1}}} \cdot \underbrace{\|(B_{s_j}-B_{s_{j-1}})^2\|_{L^2(\mathbb{P})}}_{\sqrt{3 (s_j-s_{j-1})^2}} \\ &= \sqrt{3} \cdot \sum_{j=1}^n (s_j-s_{j-1})^{\frac{3}{2}} \leq |\Pi|^{\frac{1}{2}} \cdot \underbrace{\sum_{j=1}^n (s_j-s_{j-1})}_{T} \\ &\to 0 \quad (|\Pi| \to 0) \end{align*}$$ i.e. $I_1(\Pi) \stackrel{L^1}{\to} 0$ as $|\Pi| \to 0$. Consequently, there exists a subsequence converging to $0$ almost surely.
Adding all up, we get
\begin{align*} B_T^3 &\stackrel{(1)}{=} \lim_{|\Pi| \to 0} \bigg( \sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3 + 3 \underbrace{\sum_{j=1}^n B_{s_j} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{\stackrel{(3)}{\to} \int_0^T B_s \, ds} + 3 \cdot \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}})}_{\stackrel{(2)}{\to} \int_0^T B_s^2 \, dB_s} \bigg) \\ &= 0 + 3 \cdot \int_0^T B_s \, ds + 3 \cdot \int_0^T B_s^2 \, dB_s \end{align*}
When you rewrote the differential equation as an integral equation you forgot the initial condition. Please be more careful about notation (e.g. whether it is $t$ or $T$)
If we set $f(t,x) := e^{t/2} \cos(x)$, then $Y_t := f(t,B_t)$ satisfies by Itô's formula
$$dY_t = - e^{t/2} \sin(B_t) \, dB_t.$$
This is equivalent to saying
$$Y_T \color{red}{-Y_0} =- \int_0^T e^{t/2} \sin(B_t) \, dB_t.$$
Note that $Y_0 = e^0 \cos(0)=1$. Multiplying both sides with $e^{-T/2}$, we find
$$\cos(B_T) - e^{-T/2} = - \int_0^T e^{(t-T)/2} \sin(B_t) \, dB_t.$$
Hence,
$$X_T = \cos(B_T) = e^{-T/2} - \int_0^T e^{(t-T)/2} \sin(B_t) \, dB_t.$$
This shows, in particular, $\mathbb{E}(X_T) = e^{-T/2}$.
Best Answer
Just notice that $Y_t$ is given by $Y_t \equiv e^{Z_t}$ where $Z_t$ is a solution of the SDE
$$dZ_t = h(t)dB_t - \frac12 h^2(t) dt .\tag1$$
Hence you can apply Itô's lemma to $f(t,Z_t) := e^{Z_t}$ as follows :
$$\begin{align*}dY_t = df(t,Z_t) &= f_t (t,Z_t) dt + f_x(t,Z_t)dZ_t + \frac{1}{2}h(t)^2f_{xx}(t,Z_t)dt\\ &= 0 + e^{Z_t}dZ_t + \frac{1}{2}h(t)^2 e^{Z_t}dt \end{align*}$$
Now replace $e^{Z_t}$ by $Y_t$ and $dZ_t$ by its expression given in $(1)$ and conclude.