Given,
$$
3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)
$$
$X$ is poisson distribution this gives
$$
P(X=x)=\frac{e^{-\lambda} \lambda^x}{x !}, \quad[x=0,1,2, \ldots]
$$

Now, $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$
$$
\begin{aligned}
& \Rightarrow \quad \frac{3 e^{-\lambda} \lambda^4}{4 !}=\frac{1}{2} \frac{e^{-\lambda} \lambda^2}{2 !}+\frac{e^{-\lambda} \lambda^0}{0 !} \\
& \Rightarrow \quad \frac{3 \lambda^4}{4 !}=\frac{\lambda^2}{4}+1 \\
& \Rightarrow \quad \frac{\lambda^4}{8}=\frac{\lambda^2}{4}+1 \text { or } \lambda^4-2 \lambda^2-8=0
\end{aligned}
$$

Let $\lambda^2=u$, then $u^2-2 u-8=0$
Gives, $u=4,-2$

\begin{array}{ll}
\therefore & \lambda^2=4 \\
\Rightarrow & \lambda=2
\end{array}
$\left[\because \lambda^2 \neq-2\right]$

\text { Mean of } X=2
$$