From wikipedia, the moments are given by the formula
$$E[X^n] = \theta^n \cdot \frac{\Gamma(n+k)}{\Gamma(k)}$$
where $\theta$ is the scale parameter and $k$ is the shape parameter.
Since you stated that you found the correct answer in the first part, you are not using the standard notation. Your $\alpha$ is the shape parameter and $\beta$ is the scale paarameter.
There is no well-known distribution for the weighted sum of a random vector with a Dirichlet distribution. However, as partially checked in this old answer, the beta distribution can be a good approximation for it (you do not need to normalize the vector $p$ as it is already normalized).
This 2023 paper derives a novel integral representation
for the density of a weighted sum of Dirichlet distributed random variables (Appendix A.1, page 15); you can use it if you want the exact distribution. This paper also presents various non-asymptotic Gaussian-based bounds for probabilities of linear transformations of a Dirichlet random vector.
Regarding your results: Note that for $c>0$, and $X$ and $Y$ that are independent with
$$X \sim \text{Gamma} (\alpha_1, \lambda), Y \sim \text{Gamma} (\alpha_2, \lambda),$$
we have
$$ cX \sim \text{Gamma} \left ( \alpha_1, \frac{\lambda_1}{c} \right ) $$
$$X +Y \sim \text{Gamma} \left (\alpha_1+\alpha_2, \lambda \right).$$
Hence, generally there is no $\alpha'$ and $\lambda'$ such that $$p_1X+p_2Y \sim \text{Gamma} \left (\alpha', \lambda' \right), $$ unless $p_1=p_2=p$, for which we have $$pX+pY \sim \text{Gamma} \left (\alpha_1+\alpha_2, \frac{\lambda}{p} \right).$$
Best Answer
It is not difficult. I cannot show you the entire proof because you question is without your own work but this is a useful hint:
To simplify the notation, let me set $X,Y$ as the two independent Gamma rv and let's derive the law of
$$U=\frac{X}{X+Y}$$
The starting point is the following system
$$\begin{cases} u=\frac{x}{x+y} \\ v=x \end{cases}\rightarrow \begin{cases} x=v \\ y=v\frac{1-u}{u} \end{cases}$$
with Jacobian $|J|=\frac{v}{u^2}$
Substitute in $f_{XY}(x,y)$ and solve the integral in $dv$ finding your beta density. It is not difficult so show your works amending your question and, just in case, I will take you to the solution