My solution attempt for part (C)
$$\mathbb{E}[e^{\mu+\sigma Z}I_{Z>−d}]= \int^{+\infty}_{-\infty} e^{\mu+\sigma Z}I_{Z>−d} \, dx= \int^{+\infty}_{-d} e^{\mu+\sigma Z} \, dx$$ since $$\int^{-d}_{-\infty} e^{\mu+\sigma Z} *0=0$$
So I am left to integrate $$\int^{+\infty}_{-d} e^{\mu+\sigma Z} \, dx=e^\mu\int^{+\infty}_{-d} e^{\sigma Z} \, dx$$ which I don't know how to do since $Z$ is a r.v.
By definition of expectation, \begin{align*} & E[e^{\mu + \sigma Z}I_{Z > -d}] \\ = & \int_{-d}^\infty \exp(\mu + \sigma z) \phi(z) d z \\ = & \frac{1}{\sqrt{2\pi}}\int_{-d}^\infty \exp\left(\mu + \sigma z - \frac{1}{2}z^2\right) d z \\ = & \frac{1}{\sqrt{2\pi}}\exp\left(\mu + \frac{1}{2}\sigma^2\right)\int_{-d}^\infty \exp\left(-\frac{1}{2}(z - \sigma)^2\right) d z \\ = & \exp\left(\mu + \frac{1}{2}\sigma^2\right) P\left(X > -d\right) \\ = & \exp\left(\mu + \frac{1}{2}\sigma^2\right)(1 - \Phi(-d - \sigma)) \\ = & e^{\mu + \frac{1}{2}\sigma^2} \Phi(d + \sigma), \end{align*} where $X \sim \mathcal{N}(\sigma, 1)$, in view of its pdf appeared as integrand.