Here is a solution to (1). Not sure if it strictly falls under the requirements of the problem, but it doesn't use anything too fancy.
Let $\tau_r$ be the first time Brownian motion hits either $1$ or $-r$ and $f(x)=x^2$. Since the infinitesimal generator of Brownian motion (restricted to smooth functions) is $1/2 \Delta$, we have by Dynkin's formula that
$$\mathbf{E}[f(B_{\tau_r})]=\mathbf{E}[B_{\tau_r}^2]=f(0)+\mathbf{E}\left[ \int_0^{\tau_r} 1 ds \right]=\mathbf{E}[\tau_r] .$$
Using the Gambler's ruin estimate, we have
$$\mathbf{E}[B_{\tau_r}^2]=\frac{r^2}{r+1}+\frac{r}{r+1}.$$
Letting $r \rightarrow \infty$, we see that $\mathbf{E}[\tau_r] \rightarrow \infty$ as $r \rightarrow \infty$. It follows that $\mathbf{E}[\tau]=\infty$.
Basically, you can simply apply Itô's formula:
$$\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^2 \\ &\quad +\frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad + \int_0^t f_{yy}(W_s^1,W_s^2) \, \underbrace{dW_s^2 dW_s^2}_{ds} \bigg). \end{align*}$$
If we choose $f(x,y) := \exp(\sigma(x+y))$ and take the expectation on both sides, it follows that
$$\mathbb{E}e^{\sigma (W_t^1+W_t^2)}-1 = \sigma^2(\varrho+1) \int_0^t \mathbb{E}e^{\sigma (W_s^1+W_s^2)} \, ds.$$
Setting $\varphi(t) := \mathbb{E}e^{\sigma (W_t^1+W_t^2)}$ we find that $\varphi$ is a solution to the ordinary differential equation
$$\varphi'(t) = \sigma^2 (\varrho+1) \varphi(t) \qquad \varphi(0)=1.$$
This ODE can be solved explicitly,
$$\varphi(t) = \exp(\sigma^2(\varrho+1)t).$$
Combining the results, we conclude
$$\mathbb{E}(X_t\cdot Y_t) = X_0 Y_0 e^{2\mu t} e^{\sigma^2 \varrho t}.$$
Best Answer
I'll assume that $c,d,e$ depend only on $t$. In integral form, $$ Y_t=Y_0+\int_0^tc_s\,\mathrm ds+\int_0^td_s\,\mathrm dW_s^1+\int_0^te_s\,\mathrm dW_s^2. $$ Then, for any $C^2$ function $f$ Itô's formula (in differential form) states $$ \mathrm d\left(f(Y_t)\right)=f'(Y_t)\,\mathrm dY_t+\frac12f''(Y_t)\,\mathrm d\langle Y\rangle_t. $$ I think your problem is finding $\mathrm d\langle Y\rangle_t$. To do this, use bilinearity and symmetry of $\langle \cdot,\cdot\rangle$: \begin{multline*} \langle Y\rangle_t=\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t+\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t +2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t\\ +2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t+2\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t. \end{multline*} Now, $\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t=0$ and $\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot \varphi_s\,\mathrm dW_s^i\rangle_t=0$ because $\int_0^\cdot c_s\,\mathrm ds$ has finite variation. Also, $\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t=0$ because the two Brownian motions are independent.
The only terms that remain are $$ \langle Y\rangle_t=\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t =\int_0^t d_s^2\,\mathrm ds+\int_0^t e_s^2\,\mathrm ds=\int_0^t \left(d_s^2+e_s^2\right)\,\mathrm ds. $$ Thus, $$ \mathrm d\left(f(Y_t)\right)=f'(Y_t)\left(c_t\mathrm dt+d_t\mathrm dW^1_t+e_t\mathrm dW^2_t\right)+\frac12f''(Y_t)\left(d_t^2+e_t^2\right)\,\mathrm dt. $$ Lastly, if you look around, you'll see that a general form of Itô's formula can be applied directly to $f(Y_t)=f\circ F(\int c_s\,\mathrm ds,\int d_s\,\mathrm dW_s^1,\int e_s\,\mathrm dW_s^2)$, making directly appear the quadratic variations of $W_t^1$ and $W_t^2$. The two approaches coincide.