You just have to sum over all probabilities. There are 3 possibilities you need to be concerned about:
(a) exactly 2 earthquakes in first 6 months, given that at least 2 happened in those 6 months, AND at least 2 earthquakes in next 3 months.
(b) exactly 3 earthquakes in first 6 months, given that at least 2 happened in those 6 months, AND at least 1 earthquake in next 3 months.
(c) at least 4 earthquakes in first 6 months, given that at least 2 happened in those 6 months.
The memoryless property means the AND is straight multiplication.
Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda$.
Step 1: $(X_t)_{t \geq 0}$ has almost surely increasing sample paths.
Proof: Fix $s \leq t$. Since $(X_t)_{t \geq 0}$ has stationary increments and $X_{t-s}$ is Poisson distributed, we have
$$\mathbb{P}(X_s>X_t) = \mathbb{P}(X_t-X_s < 0) = \mathbb{P}(X_{t-s}<0)=0,$$
i.e. $X_s \leq X_t$ almost surely. As $\mathbb{Q}_+$ is countable, this implies
$$\mathbb{P}(\forall q \leq r, q,r \in \mathbb{Q}_+: X_q \leq X_r)=1.$$
Since $(X_t)_{t \geq 0}$ has càdlàg sample paths, this already implies
$$\mathbb{P}(\forall s \leq t: X_s \leq X_t)= 1.$$
Step 2: $(X_t)_{t \geq 0}$ takes almost surely only integer values.
We have $\mathbb{P}(X_q \in \mathbb{N}_0)=1$ for all $q \in \mathbb{Q}_+$. Hence, $\mathbb{P}(\forall q \in \mathbb{Q}_+: X_q \in \mathbb{N}_0)=1$. As $(X_t)_{t \geq 0}$ has càdlàg sample paths, we get
$$\mathbb{P}(\forall t \geq 0: X_t \in \mathbb{N}_0)=1.$$
(Note that $\Omega \backslash \{\forall t \geq 0: X_t \in \mathbb{N}_0\} \subseteq \{\exists q \in \mathbb{Q}_+: X_q \notin \mathbb{N}_0$ and that the latter is a $\mathbb{P}$-null set.)
Step 3: $(X_t)_{t \geq 0}$ has almost surely integer-valued jump heights.
We already know from step 1 and 2 that there exists a null set $N$ such that $X_t(\omega) \in \mathbb{N}_0$ and $t \mapsto X_t(\omega)$ is non-decreasing for all $\omega \in \Omega \backslash N$. Consequently, we have
$$X_s(\omega) - X_t(\omega) \in \mathbb{N}_0$$
for any $s \geq t$ and $\omega \in \Omega \backslash N$. On the other hand, we know that the limit
$$\lim_{t \uparrow s} (X_s(\omega)-X_t(\omega)) = \Delta X_s(\omega) $$
exists. Combing both considerations yields $\Delta X_s(\omega) \in \mathbb{N}_0$. (Check that the following statement is true: If $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}_0$ and the limit $a:=\lim_n a_n$ exists, then $a \in \mathbb{N}_0$.) Since this holds for any $s \geq 0$, we get
$$\Delta X_s(\omega) \in \mathbb{N}_0 \qquad \text{for all $\omega \in \Omega \backslash N$, $s \geq 0$},$$
i.e. $$\mathbb{P}(\forall s \geq 0: \Delta X_s \in \mathbb{N}_0)=1.$$
Step 4: $(X_t)_{t \geq 0}$ has almost surely jumps of height $1$.
By step 3, it suffices to show that $$\mathbb{P}(\exists t \geq 0: \Delta X_t \geq 2)=0.$$
Since the countable union of null sets is a null set, it suffices to show
$$p(T) := \mathbb{P}(\exists t \in [0,T]: \Delta X_t \geq 2)=0$$
for all $T>0$. To this end, we first note that
\begin{align*} \Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\} &\subseteq \Omega_0 \cap \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \\ &\subseteq \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \end{align*}
for all $k \in \mathbb{N}$ where $$\Omega_0 := \{\omega; s \mapsto X_s \, \, \text{is non-decreasing}\}.$$ Using that $\mathbb{P}(\Omega_0)=1$ (by Step 1) and the fact that the increments $X_{\frac{j}{k}}-X_{\frac{j-1}{k}}$ are independent Poisson distributed random variables with parameter $\lambda/k$, we get
$$\begin{align*} p(T) &=\mathbb{P}(\Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\}) \\ &\leq \sum_{j=1}^{kT} \mathbb{P}(X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2) \\ &= kT \mathbb{P}(X_{\frac{1}{k}} \geq 2) = kT \left(1-e^{-\lambda/k} \left[1+\frac{\lambda}{k} \right]\right) \\ &= \lambda T \frac{1-e^{-\lambda/k} \left(1+\frac{\lambda}{k} \right)}{\frac{\lambda}{k}}. \end{align*}$$
Letting $k \to \infty$, we find
$$p(T) \leq \lambda T \frac{d}{dx} (-e^{-x}(1+x)) \bigg|_{x=0} = 0.$$
Best Answer
For a homogeneous Poisson point process (as a counting process $\{N(t,\omega): t\in\mathbb{R}_+, \omega\in\Omega\}$), the possible sample paths $\{N(\cdot,\omega):\omega\in\Omega\}$ are just right-continuous step functions with unit jumps, so each path is uniquely specified by its jump locations, say $\omega=(t_1,t_2,\ldots)\in\mathbb{R}_+^\infty.$ Once all these "arrival times" are specified, $N(\cdot,\omega)$ is fixed as the corresponding step function. (In other words, we can take the underlying "possibility space" $\Omega$ to be the set of all possible arrival-time sequences.)
The random variable $N(t)=N(t,\cdot)$ counts the number of arrivals in the interval $(0,t]$, so $N\left(t+h\right)-N\left(t\right)$ is the number of arrivals in $(t,t+h]$. Now, $P\left\{N\left(t+h\right)-N\left(t\right) \ge 2\right\} = o\left(h\right)$ means that this probability divided by $h$, approaches $0$ as $h\to 0$; i.e., the probability of more than one arrival in $(t,t+h]$, divided by $h$, approaches $0$ as the interval width $h\to 0.$ But this is exactly what we have if $N(t+h)-N(t)$ has a $\text{Poisson}(\lambda h)$ Distribution: $$\begin{align}&\lim_{h\to 0}{P(N(t+h)-N(t)\ge 2)\over h}\\ &=\lim_{h\to 0}{1-P(N(t+h)-N(t)=0)-P(N(t+h)-N(t)=1)\over h}\\[2ex] &=\lim_{h\to 0}{1-e^{-\lambda h}(\lambda h)^0/0!-e^{-\lambda h}(\lambda h)^1/1!\over h}\\[2ex] &=\lim_{h\to 0}{1-(1-\lambda h+o(h))-(1-\lambda h+o(h))(\lambda h)\over h}\\[2ex] &=0 \end{align}$$