The given MGF is that for a random variable following a Poisson distribution with parameter [tex]\lambda=3[/tex].
This means [tex]\mathbb E(X)=\mathbb V(X)=\lambda[/tex], and [tex]X[/tex] has PMF
[tex]f_X(x)=\begin{cases}\dfrac{3^xe^{-3}}{x!}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}[/tex]
So, the desired probability is
[tex]\mathbb P\left(\lambda-\lambda^2<X<\dfrac12(\lambda+\lambda^2)\right)=\mathbb P(0<X<\lambda)=\mathbb P(X<3)[/tex]
This is equivalent to
[tex]\displaystyle\sum_{x=0}^2\mathbb P(X=x)=\sum_{x=0}^2\frac{3^x}{x!e^3}=\frac{17}{2e^3}\approx0.4232[/tex]
