I have the following problem:
Let $N$ be the number of losses and let
$$ T[N] = \begin{cases} \$0 & \text{ if }N = 0\\ \$10,000 & \text{ if }N = 1 \\ \$17,500 & \text{ if }N = 2 \\ \$17,500 + 5000N & \text{ if }N \geq 3 \end{cases} $$
be the payment.
Then \begin{align*}E[T] &= 0\cdot P[N = 0] + 10000 \cdot P[N = 1] + 17500\cdot P[N = 2] + (17500 + N)P [N\geq 3]\\ &= 0\cdot P[N = 0] + 10000 \cdot P[N = 1] + 17500\cdot P[N = 2] + 17500P [N\geq 3] + N\cdot{}P [N\geq 3]\end{align*}
Now using $P[N = n] = \frac{e^{-2.5}\cdot2.5^n}{n!}$, I can calculate all of this except the term $N\cdot{}P [N\geq 3].$ But this looks a lot like the expected value, so perhaps there is a manipulation that I can try?
The first three terms in your calculation of $E[T]$ are correct but the last needs to be an infinite sum: $$E[T(N)] = \sum_{n=0}^\infty T(n)P(N=n) = \sum_{n=0}^2 + \sum_{n=3}^\infty $$ For this last sum plug in $T(n)=17.5 + 5(n-2)$ (working in \$thousands, and using @Michael's comment). Then the device is to add back some terms to create a 'nice' summation, and subtract off the extra terms: $$\begin{align} \sum_{n=3}^\infty T(n)P(N=n) &= \sum_{n=0}^\infty [7.5+5n]P(N=n)-\sum_{n=0}^2[7.5+5n]P(N=n)\\ &=\left(7.5 + 5E[N]\right) - \left( 7.5P(N=0)+12.5P(N=1)+17.5P(N=2)\right)\;, \end{align} $$ where we recognize that $\sum_0^\infty P(N=n)=1$ and $\sum_0^\infty nP(N=n)=E[N]$. Now combine this with the first three terms, and plug in the Poisson probabilities.
Note the formula $T(n)=17.5 + 5(n-2)$ also is true for $n=2$ so you can save a bit of effort by including $n=2$ in the infinite sum: $\sum_{n=0}^\infty=\sum_{n=0}^1 + \sum_{n=2}^\infty $.