Let $B$ be a standard brownian motion under the usual conditions.
Evaluate $$ \mathbb E\left[(B_s+B_t^2)e^{B_t}|\mathcal F_s\right]$$
I started by separating and adding $t/2$ to the exponential $$ \mathbb E\left[B_s e^{B_t-\frac{t}{2}}e^{\frac{t}{2}}|\mathcal F_s\right]+ \mathbb E\left[B_t^2e^{B_t-\frac{t}{2}}e^{\frac{t}{2}}|\mathcal F_s\right]$$ $$=B_s e^{\frac{t}{2}}\mathbb E\left[e^{B_t-\frac{t}{2}}|\mathcal F_s\right]+ e^{\frac{t}{2}}\mathbb E\left[B_t^2e^{B_t-\frac{t}{2}}|\mathcal F_s\right]$$ Since $e^{B_t-\frac{t}{2}}$ is a martingale, and by adding $t$ to the left side we get $$=B_s e^{\frac{t-s}{2}}e^{B_s}+ e^{\frac{t}{2}}\mathbb E\left[(B_t^2-t+t)e^{B_t-\frac{t}{2}}|\mathcal F_s\right]$$ $$=B_s e^{\frac{t-s}{2}}e^{B_s}+ e^{\frac{t}{2}}\mathbb E\left[(B_t^2-t)e^{B_t-\frac{t}{2}}|\mathcal F_s\right]+te^{\frac{t}{2}}\mathbb E\left[e^{B_t-\frac{t}{2}}|\mathcal F_s\right]$$ $$=B_s e^{\frac{t-s}{2}}e^{B_s}+ e^{\frac{t}{2}}\mathbb E\left[(B_t^2-t)e^{B_t-\frac{t}{2}}|\mathcal F_s\right]+te^{\frac{t}{2}}e^{B_s-\frac{s}{2}}$$
I don't know how to proceed, any corrections will be appreciated
The hard one to figure out is $\mathbb{E}[B_t^2e^{B_t}|\mathcal F_s]$, so that's the one I will show how to evaluate. It seems like you know \begin{align*} \mathbb{E}[e^{\lambda B_t}|\mathcal F_s] &= e^{\frac{\lambda^2}{2}t}\mathbb{E}[e^{\lambda B_t-\frac{\lambda^2}2 t}|\mathcal F_s] \\ &= e^{\lambda B_s + \frac{\lambda^2}{2}(t-s)}. \end{align*}
The trick is to notice that if we differentiate $e^{\lambda B_t}$ with respect to $\lambda$ twice, we get $B_t^2 e^{\lambda B_t}$, then setting $\lambda = 1$ gives what we want to take the conditional expectation of. So, assuming we can take the derivative outside the conditional expectation (we can show this is valid with the dominated convergence theorem), we get \begin{align*} \mathbb{E}[B_t^2 e^{\lambda B_t}|\mathcal F_s] &= \mathbb{E}\left[\left.\frac{d^2}{d\lambda^2} e^{\lambda B_t}\right|\mathcal F_s\right] \\ &= \frac{d^2}{d\lambda^2} \mathbb{E}[e^{\lambda B_t}|\mathcal F_s] \\ &= \frac{d^2}{d\lambda^2}(e^{\lambda B_s + \frac{\lambda^2}{2}(t-s)}) \\ &= e^{\lambda B_s + \frac{\lambda^2}{2}(t-s)}\big(t-s + (B_s + \lambda(t-s))^2\big). \end{align*} Now setting $\lambda = 1$ gives $$\mathbb{E}[B_t^2 e^{B_t}|\mathcal F_s] = e^{B_s + \frac{t-s}{2}}\big(t-s + (B_s + (t-s))^2\big).$$