(Personal note: there is an incredible amount of questions whose title consists in some permutation of the words I chose for the title, and I guess someone may have already answered. I swear I looked many of them up to see if someone did. If I missed it, I am sorry for repeating it.)
Be $X_t:=\int_0^t e^s\text d W_s$, and $Y_t:=\int_0^t X_s\text dW_s$. Find $\mathbb E\left[Y_t\right]$ and $\mathbb V\left[Y_t\right]$.
I see that $Y_t$ must be a martingale because it is just the martingale part of an Itô integral (even if the integrand is given as an Itô integral?), so I state $\mathbb E\left[Y_t\right]=0$.
That said, $\mathbb V\left[Y_t\right]= \mathbb E\left[\left(\int_0^t X_s\text dW_s\right)^2\right]= \mathbb E\left[\int_0^t X_s^2\text d s\right]$ by Itô isometry. But by Itô lemma, $$\text d X_s = e^s\text d W_s\Rightarrow\text d (X_s^2)=2X_s\text d X_s+\frac 12 e^{2s}\text ds=2\int_0^s e^u\text d W_u e^s\text d W_s+\frac 12 e^{2s}\text ds.$$
i) is the general reasoning sound? To me, everything seems to make sense, but I am absolutely not sure.
ii) How do I go past the expression for the variance?
I am assuming there is a final time horizon $T.$
For a suitable integrand $f$ (that make the integral well-defined), the Ito integral (as a process) $\int_0^t f(s)dW_s, 0 \leq t \leq T,$ is a local martingale, not necessarily a martingale. For the Ito integral to be a (square-integrable) martingale you need the integrand to satisfy $E\int_0^T |f(s)|^2 ds < \infty.$ In your case this condition is obviously satisfied since $e^s$ is a continuous function. Thus the process $X$ is a square-integrable martingale. Therefore, you have $E[Y_t]=0$ as you said.
For the variance, note that since $e^s$ is a deterministic function, we have that $\int_0^t f(s)dW_s \sim N(0, \int_0^t e^{2s}ds).$ Thus $$E[X^2_t]= \int_0^t e^{2s}ds = \frac{1}{2}(e^{2t} -1).$$ Also, by Ito isometry and Fubini's theorem $$var[Y_t]=E\left[\left(\int_0^t X_s dW_s \right)^2\right] = E\left[\int_0^t X^2_s d_s \right]=\int_0^t EX^2_s d_s=...$$