State if the Itô integral exists for $\int_0^T\mathrm{e}^{B_t^2}\,\mathrm{d}B_t$ and calculate its mean and variance.
Mean
Since $\mathrm{e}^x$ is continuous and bounded on $[0,T]$, then it is square integrable, i.e. $$\mathbb{E}\left(\int_0^T\mathrm{e}^{B_t^2}\,\mathrm{d}B_t\right)^2=\int_0^T\mathbb{E}\left(\mathrm{e}^{2B_t^2}\right)\,\mathrm{d}t<\infty.$$ Hence, $\int_0^T\mathrm{e}^{B_t^2}\,\mathrm{d}B_t$ has zero mean.
Variance
By Itô's lemma, $B_t^2=2\int_0^TB_t\,\mathrm{d}B_t+\int_0^T\,\mathrm{d}t$.
$2\int_0^TB_t\,\mathrm{d}B_t$ is an Itô integral hence it has mean 0, while $\int_0^T\,\mathrm{d}t=T$. Hence, $B_t^2\sim\mathcal{N}(0,T)$.
Using the moment generating variable of a normal random variable, $\mathbb{E}\left(\mathrm{e}^{2B_t^2}\right)=\mathrm{e}^{2t}$. The variance of $\int_0^T\mathbb{E}\left(\mathrm{e}^{2B_t^2}\right)\,\mathrm{d}t$ is then $$\int_0^T\mathrm{e}^{2t}\,\mathrm{d}t=\frac{\mathrm{e}^{2t}-1}2.$$
Thus, it is an Itô Integral.
Am I right?