Probability - Expectation calculation for a function. 

What you have doen for the fisrt part is correct and 
\[\mu_Y=\mu_X -L \int_L^{\infty}f_X(x)dx.\]
If $X$ has a normal distribution with mean $\mu_X$ and variance $\sigma_X^2$, then  
\[\mu_Y=\mu_X -L \int_L^{\infty} \frac{1}{\sigma_X\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx=\mu_X -\frac{L}{\sigma_X\sqrt{2\pi}} \int_L^{\infty} e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx.\]
Consider the function
\[g(\mu_X)=\mu_X -\frac{L}{\sigma_X\sqrt{2\pi}} \int_L^{\infty} e^{-\frac{1}{2}(\frac{X-\mu_X}{\sigma_X})^2}dx,      \mu_X \geq 0\]
Then $g$ is a continuous function on $[0,\infty)$ and 
\[\lim_{\mu_X \rightarrow \infty}g(\mu_x)=+\infty.\]
Hence $g$ is a coercive function and must have a finite minimum point in $[0, \infty)$. So there is a $\mu_x$ that minimizes $g(\mu_X).$ See the Theorem in the scond page of the following notes: https://www.math.usm.edu/lambers/mat419/lecture4.pdf