NOTE: The questions can answered independently, but it is preferable to follow through the reasoning.
Suppose $\{N_t\}$ is a Poisson process with intensity $\lambda=4$. I want to compute $$\mathbb{P}[N_{10}=3\mid N_{15}-N_5=3].$$ Now we have the following equality of events $$\{N_{10}=3\text{ and }N_{15}-N_5=3 \}=\{N_{10}=3\text{ and }N_{15}-N_{10}=N_5\}.$$
First Question: Is the event $\{N_{15}-N_{10}=N_5\}$ measurable w.r.t $\sigma(N_{15}-N_{10})?$
If the answer to the above question is yes, then the events $\{N_{10}=3\}$ and $\{N_{15}-N_{10}=N_5\}$ are independent and thus $$\mathbb{P}[N_{10}=3\text{ and }N_{15}-N_{10}=N_5]=\mathbb P[N_{10}=3]\cdot\mathbb P[N_{15}-N_{10}=N_5].$$
Now the random variables $N_5$ and $N_{15}-N_{10}$ are independent and therefore \begin{align*}\mathbb{P}[N_{15}-N_{10}=N_5]&=\sum_{n\in\mathbb{N}}\mathbb{P}[N_{15}-N_{10}=n]\cdot [N_5=n]=\sum_{n\in\mathbb{N}}\mathbb{P}[N_5=n]^2\\ &=\sum_{n\in\mathbb{N}}\left(\frac{20^k}{k!}e^{-20}\right)^2?\end{align*}
Second Question: Is there a "nice" formula for computing $\sum_{n\in\mathbb{N}}\left(\frac{20^k}{k!}e^{-20}\right)^2$?
If yes then we can now get the probability of our original event by using that $$ \mathbb{P}[N_{10}=3\mid N_{15}-N_5=3]=\frac{\Bbb P[N_{10}=3]}{\Bbb P[N_{15}-N_5=3]}\cdot\sum_{n\in\mathbb{N}}\Bbb P[N_5=n]^2.$$
Third Question: Is there a straighforward intuitive way of computing the probability of the original even?
In order to calculate: $$P\left(N_{10}=3,N_{15}-N_{5}=3\right)$$ so that on base of that you can calculate: $$P\left(N_{10}=3\mid N_{15}-N_{5}=3\right)=P\left(N_{10}=3,N_{15}-N_{5}=3\right)/P\left(N_{15}-N_{5}=3\right)$$ you should go for:
$$P\left(N_{10}=3,N_{15}-N_{5}=3\right)=\sum_{k=0}^{3}P\left(N_{5}=k,N_{10}-N_{5}=3-k,N_{15}-N_{10}=k\right)$$ where $N_{5},N_{10}-N_{5},N_{15}-N_{10}$ are iid $\sim\mathsf{Poisson}\left(5\lambda=20\right)$.
The answer to your first question is: "no".
If it would be true then $1_{N_{15}-N_{10}=N_{5}}$ would be a function that can be written as $f(N_{15}-N_{10})$ which means that the value it takes only depends on the values that are taken by $N_{15}$ and $N_{10}$. That is evidently not the case: it also depends on the value taken by $N_5$.