Consider a Poisson process (N(t))$_{t \geq 0}$ with $\lambda = 1.5$ compute the following
a) $P(N(2) = 2; N(2.5) = 3; N(3) = 6)$
b) $P(N(2.5)*N(3)=3)$
c) $P(N(2) + N(2.5) = 0)$
d) $P(N(2) + N(2.5) = 1)$
e) $P(N(2) + N(2.5) = 2)$
I got all of these incorrect and I was hoping that someone could guide me in the right direction
For a) I did
$P(N(2) = 2; N(2.5) = 3; N(3) = 6) \\=P(N(2) = 2)*P(N(2.5)-N(2) = 3 - 2)*P(N(3)-N(2.5)=6-3)\\
=P(N(2)=2)*P(N(0.5)=1)*P(N(0.5)=3)\\=\frac{3^2e^{-2}}{2!}*\frac{0.75^1e^{-0.75}}{1!}*\frac{0.75^3e^{-0.75}}{3!}$
and b) I got
$P(N(2.5)*N(3)=3)=P(N(16.875)=3)=\frac{16.875^3e^{-16.875}}{3!}$
c)
$P(N(2)+N(2.5)=0)=\frac{6.75^0e^{-6.75}}{0!}$
d)
$P(N(2)+N(2.5)=1)=\frac{6.75^1e^{-6.75}}{1!}$
e)
$P(N(2)+N(2.5)=2)=\frac{6.75^2e^{-6.75}}{2!}$
For (a), your initial reasoning is correct but then the last step has a minor computational error. Note that $$N(t) \sim \operatorname{Poisson}(\lambda t),$$ hence $$\Pr[N(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!}.$$ So in particular, for $\lambda = 1.5 = 3/2$, $$\Pr[N(2) = 2] = e^{-3} \frac{3^2}{2!}$$ but you have written $e^{-2}$ instead of $e^{-3}$. Similarly, we have $$\Pr[N(0.5) = 1] = e^{-3/4} \frac{(3/4)^1}{1!}$$ which you have correct, and $$\Pr[N(0.5) = 3] = e^{-3/4} \frac{(3/4)^3}{3!},$$ which you also have correct.
For (b), your reasoning is incorrect. The event $$N(2.5)N(3) = 3$$ means that the product of the total number of events observed at times $t = 2.5$ and $t = 3$ equals $3$. This can occur in only one way, because $N(t)$ is always a nonnegative integer. In particular, we must have $N(2.5) = 1$ and $N(3) = 3$. There is no other nonnegative integer factorization of $3$ that works; notice that we cannot have $N(2.5) = 3$ and $N(3) = 1$, since we require that $N(3) \ge N(2.5)$ for a single observed process. Consequently, $$\Pr[N(2.5)N(3) = 3] = \Pr[N(2.5) = 1 \cap N(3) = 3],$$ and you can evaluate this in a similar manner as you did in part (a).
For part (c), you have a misunderstanding. The sum of the variables $N(2) + N(2.5)$ is not equal to $N(4.5)$, because $N(2)$ and $N(2.5)$ do not have independent increments--they share the time interval $t \in [0, 2]$. So if we observe $N(2) = 1$, for example, then we must have $N(2.5) \ge 1$. What you did in (a) was to effectively reformulate the joint event in terms of variables over independent increments, thereby allowing you to use the product rule for the joint probability of independent events. To answer (c), you should simply observe that if $N(2.5) = 0$, then we must also have $N(2) = 0$. And the only way that their sum equals zero is if both are zero. So $$\Pr[N(2) + N(2.5) = 0] = \Pr[N(2.5) = 0].$$
For (d) and (e), the same issue applies as for part (c), except now we cannot assert, for instance, that $\Pr[N(2) + N(2.5) = 2] = \Pr[N(2.5) = 2]$, because we could have $N(2) = 1$ and $N(2.5) = 1$, or we could have $N(2) = 0$ and $N(2.5) = 2$. And this is what you need to consider for part (e). What about (d)? In what ways can $N(2) + N(2.5) = 1$?