It might be tempting to reason as follows: Since $(X_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $\mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 \int_0^t W_s \,dW_s + t.$$ As $\mathbb{E}(X_t)=0$ we thus find
\begin{align*} \mathbb{E}(W_t^2 X_t)& =2 \mathbb{E} \left( X_t \int_0^t W_s \, dW_s \right) \\ &=2 \mathbb{E} \left(\left[ \int_0^t \text{sgn}(W_s) \, dW_s \right] \left[ \int_0^t W_s \, dW_s \right] \right) \end{align*}
Applying Itô's isometry we obtain that
$$\mathbb{E}(W_t^2 X_t) =2 \mathbb{E} \left( \int_0^t W_s \, \text{sgn}(W_s) \, ds \right) =2\int_0^t \mathbb{E}(|W_s|) \, ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s \sim N(0,s)$ entails $\mathbb{E}(|W_s|) = \sqrt{(2s)/\pi}$). As $\mathbb{E}(X_t)=0$ this shows that $$\mathbb{E}(W_t^2 X_t) \neq \mathbb{E}(W_t^2) \mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
The key here is to note that the Brownian motion at time $t$ is distributed normally, with mean zero and variance (not standard deviation) $t$. To show that $\mathbb{E}[W_t] = 0$, you don't even need the fact about the variance. Just note that its mean is always zero.
To calculate $\mathbb{E}[W_t]$, note that you can first take $e^t$ out of the expectation (it is not a random variable). It remains to calculate $\mathbb{E}[e^{W_t}] = M(1)$, where $M$ is the moment generating function (MGF) of the random variable $W_t$. (You may recall that the definition of the MGF $M(\alpha)$ for a random variable $X$ is $\mathbb{E}[e^{\alpha X}]$.) We can again use the fact that $W_t$ is normally distributed. The MGF of a normal random variable is well-known:
$$M(\mathcal{N}(\mu, \sigma^2)) = e^{\mu \alpha + \alpha^2 \sigma^2 / 2}$$
By taking $\mu = 0$, $\sigma^2 = t$, and $\alpha = 1$, we get that $M(W_t) = e^{t/2}$. Hence, $\mathbb{E}[e^{t + W_t}] = e^{3t / 2}$.
Best Answer
For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$.
But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ are not (for example, $E(X_2X_1)=0$ and $E(W_2W_1)=1$).