Since \(X_1,X_2\) are independent therefore there are two tosses of a coin to generate \(X_1,X_2\) and corresponding to each rv there is an independent poisson trial. So we have the following pmf \( p_{X_1}(x_1) = \begin{cases} 0.5 & \text{for } x_1=0 \\ 0.5e^{-1}\dfrac{1}{\left (\frac{x_1-1}{2}\right )!} & \text{for } x_1 \in \{1+2k: k\in \mathbb N \cup \{0\}\} \\ 0 & \text{elsewhere} \end{cases}\) \(p_{X_2}(x_2) = \begin{cases} 0.5 & \text{for } x_2=0 \\ 0.5e^{-0.5}\dfrac{0.5^{\left (\frac{x_2-2}{3}\right )}}{\left (\frac{x_2-2}{3}\right )!} & \text{for } x_2 \in \{2+3k: k\in \mathbb N \cup \{0\}\} \\ 0 & \text{elsewhere} \end{cases}\) The required probability is then, \begin{align*} \Pr(X_1+X_2 > 5) &= 1- \Pr(X_1+X_2\le 5) \\ &= 1-(\Pr(X_1+X_2=0) + \Pr(X_1+X_2=1) + \Pr(X_1+X_2=2)\\ &+ \Pr(X_1+X_2=3) + \Pr(X_1+X_2=5) ) \end{align*}
