A stronger fact holds: If $\lambda_1\le\lambda_2\le\dots\le\lambda_m$ are the eigenvalues of $A$, and $\mu_1\le\mu_2\le\dots\le\mu_p$ are eigenvalues of $Z^TAZ$, then $\mu_k\ge\lambda_k$ for all $k=1,\dots,p$. **Update:** This fact is well-known as [Cauchy interlacing theorem][1].

**Proof.** Let $Q$ be the quadratic form on $\mathbb R^m$ defined by $A$ (that is, $Q(x)=x^TAx$ for all $x\in\mathbb R^m$), $L:\mathbb R^p\to\mathbb R^m$ the (isometric) linear map defined by $Z$. Then $Z^TAZ$ is the matrix of the quadratic form $Q'$ on $\mathbb R^p$ given by $Q'(x)=Q(L(x))$. I suggest you think of $Q'$ as the restriction of $Q$ to the subspace $L(\mathbb R^p)$ of $\mathbb R^m$.

Suppose that $\mu_k<\lambda_k$ for some $k$. Let $V$ be the $k$-dimensional subspace of $\mathbb R^p$ spanned by the first $k$ eigenvectors of $Q'$. Then $Q'(x)\le \mu_k |x|^2$ for all $x\in V$. Let $W$ be the $(m-k+1)$-dimensional subspace of $\mathbb R^m$ spanned by the eigenvectors corresponding to $\lambda_k,\lambda_{k+1},\dots,\lambda_m$. Then $Q(x)\ge\lambda_k|x|^2$ for all $x\in W$. The subspaces $W$ and $L(V)$ have nonzero intersection since the sum of their dimensions is greater than $m$. Hence there exists a nonzro vector $x\in V$ such that $L(x)\in W$. For this vector, we have $Q'(x)\le \mu_k |x|^2 <\lambda_k|x|^2$ but $Q'(x)=Q(L(x))\ge \lambda_k|L(x)|^2=\lambda_k|x|^2$, a contradiction.
Q.E.D.

Similarly (inverting all inequalities in the argument) one sees that the $k$-th largest eigenvalue of $Q'$ is no greater than the $k$-th largest eigenvalue of $Q$, answering your second question.


  [1]: http://en.wikipedia.org/wiki/Min-max_theorem#Cauchy_interlacing_theorem