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 $\mathbf{E}[\tau]=\infty$.
First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write
$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$
(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)
Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have
$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$
Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral
$$M_t := \int_0^t W_s^3 \, dW_s$$
is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields
$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$
by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.
Best Answer
If (as is customary) the stochastic integral is understood to be continuous in $t$ (a.s.) then the equality holds for all $t$ simultaneously, with probability 1. As such, $t$ can be replaced throughout by any non-negative random variable and the a.s. equality will persist.