Let $X_1\sim \text{Bin}(n_1,p)$ and $X_2\sim \text{Bin}(n_2,p)$ independent.
a)Find the distribution of $X_1|X_1+X_2=m$
b)Calculate $E[X_1|X_1+X_2=m]$
I know that $X_1+X_2\sim \text{Bin}(n_1+n_2,p)$ \begin{align*} P(X_1|X_1+X_2=m)&\overset{\text{indep}}=\frac{P(X_1=x_1P(X_2=m-x_1)}{P(X_1+X_2=m)}\\ &=\frac{\binom{n_1}{x_1}p^{x_1}(1-p)^{n_1-x_1}\binom {n_2}{x_1-m}p^{m-x_1}(1-p)^{n_2-m+x_1}}{\binom{n_1+n_2}{m}p^m(1-p)^{n_1+n_2-m}}\\ &=\frac{\binom{n_1}{x_1}\binom{n_2}{x_1-m}}{\binom{n_1+n_2}{m}}\\ &\sim\text{Hypergeometric}(n_1+n_2,n_1,m) \end{align*}
b)I know that $$E[X_1|X_1+X_2=m]=\sum_{x}x P(X_1|X_1+X_2=m)$$
but since that I have the conditional distribution, this expectation not would be $$\frac{m\cdot n_1}{n_1+n_2}$$
Forgive me for any mistake, been a while since I study probability
$\checkmark$ Mostly okay. Excellent work; but presentation needs care.
Typo alert: That should be: $\mathsf P(X_1\color{blue}{=x_1}\mid X_1+X_2=m) ~=~ \dfrac{\dbinom{n_1}{x_1}\dbinom{n_2}{\color{blue}{m-x_1}}}{\dbinom{n_1+n_2}{m}}$.
However, that is indeed the Hypergeometric Distribution; which is that of the count of successes in a sample of size $n_1$, drawn without bias from a population of size $n_1+n_2$, which contains $m$ successes.
Thus the expectation is known to be:
$$\mathsf E(X_1\mid X_1+X_2=m) = \dfrac{n_1\cdot m}{(n_1+n_2)}$$
PS: You should include the support of $X_1$.