Below is my solution for part a, but I don't know how to solve part b. I am also unsure of my part a solution is correct or not.
a)
Since $X$ is exponentially distributed, $f_X(x) = \lambda e^{-\lambda x}$, for $x \geq 0$
Also we know that, $f_X(x) = 0$, for $x \lt 0$
$F_Y(y) = \frac{\lambda}{M(c)} \int_{x = 0}^{y} e^{cx} e^{-\lambda x} dx = \frac{\lambda}{M(c)} \int_{x = 0}^{y} e^{(c- \lambda)x} dx$
$F_Y(y) = \frac{\lambda}{(c-\lambda)M(c)} e^{(c-\lambda)y} - \frac{\lambda}{(c-\lambda)M(c)}$
So by definition, $f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{\lambda}{M(c)} e^{(c-\lambda)y}$
Where $M(c) = \frac{\lambda}{\lambda - c}$,as we plug this into the equation
$f_Y(y) = (\lambda - c) e^{(c-\lambda)y}$
b)
By definition
$M_Y(y) = E[e^{cY}] = (\lambda - c) \int e^{cy} e^{(c-\lambda)y} dy$