Answer
$\color{blue}{\frac{53}{512} \;\approx\; 0.104}$
Work Step by Step
Since $M_X(t) = (\frac{1}{4}+\frac{3}{4}e^t)^5$, then $X\sim\text{Binomial}(n=5,p=\frac{3}{4})$.
It follows that
$\qquad \displaystyle f_X(x) = {5\choose x}\left(\dfrac{3}{4}\right)^x\left(\dfrac{1}{4}\right)^{5-x},\ x=0,1,2,3,\ldots.$
Thus,
$\begin{align*}
P(X\le 2) &= P(X=0) + P(X=1) + P(X=2) \\
&= f_X(0) + f_X(1) +f_X(2) \\
&= {5\choose 0}\left(\dfrac{3}{4}\right)^0\left(\dfrac{1}{4}\right)^5 + {5\choose 1}\left(\dfrac{3}{4}\right)^1\left(\dfrac{1}{4}\right)^4 + {5\choose 2}\left(\dfrac{3}{4}\right)^2\left(\dfrac{1}{4}\right)^3 \\
&= \frac{1}{4^5} + \frac{5(3)}{4^5} + (10)\frac{9}{4^5} \\
&= \frac{106}{4^5} \\
\color{blue}{P(X\le 2)}\ &\color{blue}{= \frac{53}{512} \;\approx\; 0.104}
\end{align*}$
