Sum of two density matrices: $\rho=p_1\rho_1+p_2\rho_2$
Hints:
A density operator is by definition a (semi-)positive operators with trace equal to one.
OP is essentially asking
Is a convex linear combination of density operators again a density operator?
Answer: Yes, because:
Semi-positive operators form a cone, and
trace is linear.
To see pt. 1, note that operators $A$ in complex$^1$ Hilbert spaces enjoy the characterizations
$$A ~\text{semi-positive}\qquad \Leftrightarrow \qquad \forall v\in H~:~ \langle v| Av \rangle ~\geq~ 0, $$
and
$$A ~\text{Hermitian} \qquad \Leftrightarrow \qquad \forall v,w\in H~:~ \langle v| Aw \rangle ~=~\langle Av| w \rangle $$ $$\qquad \Leftrightarrow \qquad \forall v\in H~:~ \langle v| Av \rangle ~=~\langle Av| v \rangle \qquad \Leftrightarrow \qquad \forall v\in H~:~ \langle v| Av \rangle ~\in \mathbb{R}. $$
--
$^1$These characterizations do not hold for real Hilbert spaces, so here we are using that quantum mechanics are formulated in complex Hilbert spaces.
That depends on what you understand "density matrix" to mean. You seem to think that it refers to operators of the form $$\rho=\sum_{k=1}^n p_k|\psi_k\rangle\langle\psi_k|,\tag{1}$$ where $p_k\geq0$ for all $k$, $\sum_{k=1}^n p_k=1$, and the $|\psi_k\rangle$ are vectors in some $N$-dimensional Hilbert space $\mathcal{H}$. As a contrast, QMechanic's answer relies on a characterization of density matrices as positive semidefinite hermitian operators with trace 1. Proving the equivalence of these definitions is a very informing exercise and it will probably teach you more than your current problem.
To prove that $\rho=p_1\rho_1+p_2\rho_2$ is a density matrix by your definition, you will need to rely on an eigenvalue-eigenvector decomposition. Writing $\rho$ as in equation (1) is possible because $\rho$ is hermitian; the problem is then proving the two conditions on the $p_k$. The first one is equivalent to $\rho$ being positive semidefinite (why?), and this you can prove using the (real) abstract definition of that: $$\langle v|\rho v\rangle\geq0\,\forall v\in \mathcal{H}.$$ The sum condition you can prove by taking the trace of the different expressions you have for $\rho$.