The reason is that the compounded return of an investment has a "drag" when the intermediate returns are more volatile. Check for instance the compounded return of two return series with the same arithmetic mean, but different variance. You will see that the series with a lower variance has a higher compounded return (and geometric mean). The reason is that if an investment looses 50% of its value, it has to make a return of +100% (very different from +50%) to come back to the initial value.
The value of a bank account with no risk paying a continuously compounded interest rate $r$ is $S_0e^{r t}$. However, the GBM $S(t)$ is a model for the value of a risky/stochastic investment.
As the exact solution for $S(t)$ shows, the correct interpretation of the value of $S(t)$ as paying a continuously compounded rate of return is with $\gamma = r - \frac{1}{2}\sigma^2$ (also called growth rate) and not $r$, because $$\underset{t\rightarrow \infty}{\lim} S(t)=S_0e^{\gamma t}$$ Although the mean of $S(t)$ is $E[S(t)]=S_0e^{r t}$, the median of its distribution is $S_0e^{\gamma t}$.
While the intuition of an expected value as a proxy for the long-term value is appropriate for random variables with broadly symmetric distributions, such as returns, $dS(t)/S(t)$ (and other stationary series), in the case of the value of $S(t)$ this intuition is flawed.
To see this, notice that if $\gamma>0$ then $\underset{t\rightarrow \infty}{\lim} S(t)=\infty$, and if $\gamma<0$ then $\underset{t\rightarrow \infty}{\lim} S(t)=0$. On the other hand, $r>0$ does not guarantee an increasing value of $S$ in the long run, because if $\sigma$ is large enough, $\gamma$ could be negative even if $r$ is positive. Hence, the growth rate $\gamma$ is a better indicator of the long-term value of an investment than $r$.
Another way to see this is to notice that if $\gamma>0$, as $t$ increases the distribution of $S(t)$ becomes more skewed to the right with an ever greater right tail. As a consequence, as time passes the mean of the distribution moves deeper into the right tail, exponentially diverging from the center of the distribution, i.e. the median. To see this, notice that the ratio of the mean to the median is equal to $e^{\frac{1}{2}\sigma t}$.
Notice that the compounded rate of return $\gamma$ and the instantaneous rate of return $r$ $$\frac{d S(t)}{S(t)} = r t + \sigma dW_t$$ coincide for the value of a bank account with no risk paying interest rate $r$, i.e. $r = \gamma$ because in such case there is no randomness (aka risk) and hence $\sigma =0$. But for a risky investment, the compounded rate of return $\gamma < r$!
There is no contradiction between your two approaches. If we define
$$X_t := S_0^2 \exp \left( (2 \mu - \sigma^2)t +2 \sigma W_t \right)$$
(which is the process which you got in your first solution) then a straight-forward application of Itô's formula shows that
$$dX_t = 2 \sigma X_t \, d W_t + \left( 2\mu + \sigma^2 \right) X_t \, dt,$$
and this is exactly the SDE which you got in your second approach.
To get from the second solution to the first one you may note that
$$d(S_t^2) = (2\mu+\sigma^2) S_t^2 \, dt + 2 \sigma S_t^2 \, dW_t$$
shows that $X_t := S_t^2$ solves the SDE
$$dX_t = (2\mu+\sigma^2) X_t \, dt + 2 \sigma X_t \, dW_t,$$
and as you pointed out in your first approach the solution to this SDE is given by
$$X_t = X_0 \exp \left((2\mu+\sigma^2-(2\sigma)^2/2)t + (2\sigma) W_t \right)$$
which is exactly the process which you got in your first solution.
Best Answer
Assuming $T$ is just a number and not a stopping time, $$ E\int_0^T S_t e^{r(T-t)}dt $$ is just just a double integral of a positive function, so we can exchange the order of integration $$ = \int_0^T E [S_t] e^{r(T-t)} dt $$ $$ = \int_0^T S_0 e^{\mu t} e^{r(T-t)}dt $$ and I presume you can finish from here as nothing left is random.