How do this question? A discrete random variable $X$ can only take the value of 1, 2, 3 and 4, with the probabilities $P(X=1)=P(X=3)$ and $P(X=2)=P(X=4)=2P(X=1)$.
A random sample of size 3 is chosen from $X$ with replacement.
(a) given that $X_1,\ X_2,\text{ and }X_3$ are the three independent random values of $X$, find the probability that exactly one of these values is more than 3. $\left[ans:\dfrac{4}{9}\right]$
(b) find the probability that the sample mean is greater than  $\dfrac{4}{3}$. $\left[ans:\dfrac{209}{216}\right]$

I do part (a) like this,
$\Sigma P(X=x)=1$
$P(X=1)+P(X=2)+P(X=3)+P(X=4)=1$
$P(X=1)+2P(X=1)+P(X=1)+2P(X=1)=1$
$6P(X=1)=1$
$P(X=1)=\dfrac{1}{6}$




$X$
$1$
$2$
$3$
$4$




$P(X=x)$
$\dfrac{1}{6}$
$\dfrac{2}{6}$
$\dfrac{1}{6}$
$\dfrac{2}{6}$




$\dfrac{\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{1}{6}\right)^2 \cdot \dfrac{2}{6} \cdot 3!}{2!}=\dfrac{4}{9}$

My part (b),
$E(X)=\Sigma xP(X=x)=\dfrac{8}{3}$
$E(X^2)=\Sigma x^2P(X=x)=\dfrac{25}{3}$
$Var(X)=\sigma^2=E(X^2)-[E(X)]^2=\dfrac{17}{3}$
$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{\dfrac{17}{3}}{3}\right)$
$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{17}{9}\right)$
$P(X>\dfrac{4}{3})=P\left(\dfrac{\overline X-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}>\dfrac{\dfrac{4}{3}-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}\right)$
$=P(Z>-0.5601) $
$=1-P(Z<-0.5601) $
$=1-0.2877 $
$=0.7123 \neq \dfrac{209}{216}$
Can anyone check my work?

Your calculations for (a) is correct.  You choosing three events from




event
probability




$\{X \le 3 \}$
$2/3$


$\{X > 3 \}$
$1/3$




You need $2!$ in the denominator as events with two '$\le3$' like $\{X_1\le3, X_2\le3, X_3>3\}$ are double-counted by $3!$.

Your error in (b) comes from $\bar{X} \sim \mathcal{N}(E[X], \frac13 \mathrm{Var}(X))$.  Why does the sample mean $\bar{X}$ follow the normal distribution?
As @sudeep5221 points out, the required probability is
$$\begin{aligned}
P(\bar{X} > 4/3) &= P(X_1+X_2+X_3 > 4) \\
&= 1 - P(X_1+X_2+X_3 \le 4) \\
&= 1 - P(X_1=X_2=X_3=1) - P(X_1+X_2+X_3 = 4) \\
&= 1 - (1/6)^3 - 3 \cdot (1/6)^2 (2/6) \\
&= \frac{6^3 - 1 - 3 \cdot 1 \cdot 2}{6^3} \\
&= \frac{209}{216}
\end{aligned}$$
