Let me expand on mpiktas answer. As you noted, $W_t$ and $\int_0^t {W_s ds}$ are zero mean normal with variance $t$ and $t^3/3$, respectively (the first is trivial). Now, if $(W_t,\int_0^t {W_s ds})$ is bivariate normal, then any linear combination $aW_t + b\int_0^t {W_s ds}$, $a,b \in \mathbb{R}$, is univariate normal. In particular, the sum $W_t + \int_0^t {W_s ds}$ is (zero mean) normal. Since you are given the variance of $W_t$ and $\int_0^t {W_s ds}$, the variance of the sum follows straight from the covariance ${\rm Cov}(W_t,\int_0^t {W_s ds})$, that is from ${\rm E}[W_t \int_0^t {W_s ds}]$. Now, assuming you found that variance, calculating ${\rm E}[\exp (W_t + \int_0^t {W_s ds} )]$ amounts to calculating the moment-generating function at $1$ of a ${\rm N}(0,\sigma^2)$ variable.
EDIT: Regarding the first question, let's explain why we should expect $(W_t,\int_0^t {W_s ds})$ to be (bivariate) normal. Since $W$ is continuous on the interval $[0,t]$, $\int_0^t {W_s ds}$ is a standard Riemann integral, and $\int_0^t {W_s ds} \approx \frac{t}{n}\sum\nolimits_{k = 0}^{n - 1} {W_{kt/n} }$. Denote the right-hand side by $Y_t^n$. Then $(W_t,Y_t^n)$ is normal. Indeed, $W$ is a Gaussian process; hence, by definition, $(W_{t_1},\ldots,W_{t_m})$ is normal for any choice of times $t_1,\ldots,t_m$. But $(W_{t_1},\ldots,W_{t_m})$ is normal if and only if any linear combination $\sum\nolimits_{i = 1}^m {a_i W_{t_i } }$, $a_i \in \mathbb{R}$, is (univariate) normal, and the conclusion follows. Finally, since $(W_t,Y_t^n)$ is normal, we should expect that in the limit as $n \to \infty$, $(W_t,\int_0^t {W_s ds})$ is normal.