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 $\mathbb{E}(W_t^4) = 3t^2$.
I think I finally found a very simple and nice solution, which uses almost no calculation. Here it is.
First, we can write $W_2 = (W_2-W_1) + W_1$, and if we write $X=W_2-W_1$, $Y=W_1$, from the data of the problem we see that $X$ and $Y$ are both standard normal random variables, and independent with resepct to each other. Therefore, the problem asks for the computation of the conditional probability
$$\mathbb P(X+Y<0 | Y>0) = \frac{\mathbb P(X+Y<0\cap Y>0)}{\mathbb P(Y>0)}.$$
Now, ${\mathbb P(Y>0)}=1/2$ is immediate since $Y$ is standard normal. To compute the probability $\mathbb P(X+Y<0\cap Y>0)$, let us argue as follows: consider the joint density of $(X, Y)$ on a cartesian diagram labeled $(X,Y)$. Therefore, we are asking for the measure of the set obtained as the intersection of the sets $(X+Y<0\cap Y>0)$. This set corresponds to the lower half of the IInd quadrant on the Cartesian plane, and its measure is thus $1/8$. It follows that the measure of $\mathbb P(X+Y<0 | Y>0)$ is $\frac{1/8}{1/2}$, which is thus equal to $1/4$.
Best Answer
You can extend your first two answers readily to the third. Consider that for $t\in[0,1]$ we know that $\mathbb{E}(W_t|W_1) = t\, W_1$ but this is really $\mathbb{E}(W_t|W_1,W_0) = t\,W_1 + (1-t)W_0$ just that $W_0=0$. In general we have
$$\mathbb{E}(W_t|W_r,W_s) = \frac{t-r}{s-r} W_s + \frac{s-t}{s-r} W_r \qquad\text{for }\,\, r\leq t\leq s$$
This is the general result for a Brownian bridge. We can combine the $\min$ and $\max$ functions ($\wedge$ and $\vee$ respectively) to produce an arbitrary continuous piecewise linear function. Specifically in our case we get
$$\mathbb{E}(W_t|W_2,W_1,W_0=0) = (t\wedge1)W_1 + (0\vee(1\wedge t))(W_2-W_1)$$
For the Brownian bridge we also have that
$$\mathrm{Var}(W_t|W_r,W_s) = \frac{(s-t)(t-r)}{s-r} \qquad\text{for }\,\, r\leq t\leq s$$
Thus we have
$$\begin{align}\mathrm{Var}(W_t|W_1,W_0) &= t(1-t) \qquad&\text{for }\,\, 0 \leq t < 1\\ \mathrm{Var}(W_t|W_2,W_1)&= (2-t)(t-1) \qquad&\text{for }\,\, 1 \leq t < 2 \end{align}$$
These are parabolas, which are negative outside the intervals $[0,1]$ and $[1,2]$ respectively thus we can combine them easily with the $\max$ function as well as accounting for the linearly increasing variance for $t>2$ giving
$$\mathrm{Var}(W_t|W_2,W_1,W_0=0) = \big(t(1-t)\big)\vee\big((2-t)(t-1)\big)\vee(t-2)$$
Now, exactly as you did for your second answer, we combine the variance and square of the expectation through $$\mathbb{E}(W_t^2|W_2,W_1,W_0=0) = \mathrm{Var}(W_t|W_2,W_1,W_0=0) + \mathbb{E}(W_t|W_2,W_1,W_0=0)^2,$$ to get the final answer. (I haven't actually written out in full but you should be able to see what it looks like and if it simplifies.)