Basically the question is in two parts:
$1.)$Finding the characteristic of $P{(\lambda)}$, and it is given, I just do not know how to get the sum that they got in the very last step in this expression(last equality is unclear): $$f(t)=\sum_{j=0}^{\infty}e^{itj}\frac{\lambda^j}{j!}e^\lambda=e^{\lambda(e^{it}-1)}$$
$2.)$Finding the characteristic function of $N(0,1)$ and the last equality in the expression is unclear.
$$ f_X(t) = \int_{\mathbb R}\frac{e^{itx}}{\sqrt{2 \pi}}e^{-x^2/2}dx = \frac{1}{2\pi}\int_{\mathbb R}\cos(tx)e^{-x^2/2}dx $$
(I thought the reason for this is that the product of a even and odd function in an integral is $0$ but that's just probably my wishful thinking doing )
and the very last part that is unclear is that is says after differentiating this last line we get: $$ f_{X}'(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} (-x)\sin (tx)e^{-x^2/2}dx $$
Your expression gives $$ \begin{split} f(t) &= \sum_{j=0}^\infty e^{itj} \frac{\lambda^j}{j!}e^\lambda \\ &= e^\lambda \sum_{j=0}^\infty \frac{\left(e^{it} \lambda\right)^j}{j!} \\ &= e^\lambda e^{\lambda e^{it}} \\ &= e^{\lambda(e^{it}+1)} \end{split} $$
You likely meant $e^{-\lambda}$ not $e^\lambda$: $$ \begin{split} f(t) &= \sum_{j=0}^\infty e^{itj} \frac{\lambda^j}{j!}e^{-\lambda} \\ &= e^{-\lambda} \sum_{j=0}^\infty \frac{\left(e^{it} \lambda\right)^j}{j!} \\ &= e^{-\lambda} e^{\lambda e^{it}} \\ &= e^{\lambda(e^{it}-1)} \end{split} $$
As for (2), recall Euler's formula, $$ e^{ix} = \cos(x) + i \sin(x), \quad i = \sqrt{-1} $$