Let $\lambda_i$ be the eigenvalues of $A$ and $\beta_i$ be the eigenvalues of $B$, both matrices are symmetric of the same order $n$.
Is this equivalence true?
$$\max(\lambda_i) < \min(\beta_i) \iff A \prec B$$
Is there an analogous equivalences for positive semidefinite, negative definite and negative semidefinite?
One directions is true, the other is not:
1. It's true that $\lambda_{max}(A) < \lambda_{min}(B) \Rightarrow A \prec B$:
True, because $x^*Ax \leq \lambda_{max}(A) \left\| x \right\|^2_2 < \lambda_{min}(B) \left\| x \right\|^2_2 \leq x^*Bx$, hence $x^*Ax < x^*Bx \Leftrightarrow x^*(A-B)x <0 \Leftrightarrow A\prec B$, where we used $\lambda_{min} (M)\left\| x \right\|^2_2 \leq x^*Mx \leq \lambda_{max} (M)\left\| x \right\|^2_2 $.
Proof of $\lambda_{min} (M)\left\| x \right\|^2_2 \leq x^*Mx \leq \lambda_{max} (M)\left\| x \right\|^2_2 $:
Because $M$ is symmetric, we have $M = U^*\Lambda U$ with $U$ orthogonal and $\Lambda$ diagonal. Hence, $x^*Mx = x^*U^*\Lambda Ux$ and defining $z = Ux$, notice that since $x^*Mx = \sum^{n}_{i=1} \Lambda_{ii} z^2_i$ we got $\lambda_{min} (M)\left\| x \right\|^2_2=(\min_{i} \Lambda_{ii} ) \sum^{n}_{i=1} z^2_i \leq \sum^{n}_{i=1} \Lambda_{ii} z^2_i=x^*Mx\leq (\max_{i} \Lambda_{ii} ) \sum^{n}_{i=1} z^2_i = \lambda_{max} (M)\left\| x \right\|^2_2$, where $\sum^{n}_{i=1} z^2_i = \left\|x \right\|^2_2$, because $U$ is orthogonal.
2. But $A \prec B$ does not necessarily imply $\lambda_{max}(A) < \lambda_{min}(B)$:
Consider the counterexample $A = \begin{bmatrix} 1 & 0&0\\0&2&0\\0&0&3 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 0&0\\0&3&0\\0&0&4 \end{bmatrix}$. Here $A-B = -I_3$, but $\lambda_{max}(A) = 3 > 2 = \lambda_{min}(B)$
This analysis can be extended to each of the other "definiteness cases".