With $X\sim\operatorname{Poisson}(2)$, $Y\sim\operatorname{Poisson}(3)$ being statistically independent where $T=X+Y$ the problem asks to determine:
$1.$$P(X=k\mid T=n)$ for every value of $k$ and $n$ and obtain $E(X\mid T)$ and $\sigma (X\mid T)$
$2.$$P(T=n\mid X=k)$ for every value of $k$ and $n$ and obtain $E(T\mid X)$ and $\sigma (T\mid |X)$
To find $P(X=k\mid T=n)$ I get:
\begin{align*} P(X=k\mid T=n) &= \frac{P(X=k)\cap P(T=n)}{P(T=n)} \\ &= \frac{P(X=k)\cap P(X+Y=n)}{P(T=n)}) \\ &= \frac{P(X=k)\cap P(Y=n-k)}{P(T=n)} \end{align*}
but from here I'm not sure how the textbook concluded that this equals: \begin{align} \binom{n}{k} (\frac{2}{5})^k (\frac{3}{5})^{n-k} = X\mid T\sim\operatorname{binomial}(T,2/5) \end{align} and but I get how it follows to find the expected value and $\sigma$.
And to find $P(T=n | X=k)$ I get:
\begin{align*} P(X=k | T=n) &= \frac{P(X=k)\cap P(T=n)}{P(X=k)} \\ &= \frac{P(X=k)\cap P(Y=n-k)}{P(X=k)} \\ &= P(Y=n-k) \\ &= e^{-3}\frac{3^{n-k}}{(n-k)!} \end{align*}
but again, I'm confused to how this is equal to:
\begin{align*} T-k&\mid X=k \sim\operatorname{Poisson}(3) \\ T-X&\mid X \sim\operatorname{Poisson}(3) \end{align*}
Let $X,Y$ be independent Poisson variates with parameters $\lambda_1,\lambda_2$. Then $T=X+Y$ is a Poisson variate with parameter $\lambda_1+\lambda_2$. This is because the $MGF$ of $T$ is$$M_T(\theta)=E[e^{\theta(X+Y)}]=M_X(\theta)\cdot M_Y(\theta)=e^{(\lambda_1+\lambda_2)(e^\theta-1)}$$So in your case,$$\frac{P(X=k)\cdot P(Y=n-k)}{P(T=n)}=\frac{\left(\frac{2^ke^{-2}}{k!}\right)\cdot\left(\frac{3^{n-k}e^{-3}}{(n-k)!}\right)}{\left(\frac{5^ne^{-5}}{n!}\right)}=\binom nk\left(\frac25\right)^k\left(\frac35\right)^{n-k}$$For your second question,$$P(T=n|X=k)=P(T-X=n-k|X=k)=e^{-3}\cdot\frac{3^{n-k}}{(n-k)!}$$ $\forall n\ge k$, and hence$$P(T-X=m|X=k)=e^{-3}\cdot\frac{3^m}{m!}~\forall m\in\Bbb Z_{\ge0}$$indicating that $T-X|X$ is a Poisson variate with $\lambda=3$.