Positive sectional curvature does not imply positive definite curvature operator?
For the first question: positive curvature operator on a compact manifold implies that the manifold is diffeomorphic to a space form, i.e., a manifold of sectional curvature one. This is due to C. Boehm and B. Wilking Manifolds with positive curvature operators are space forms Annals of Mathematics, 167 (2008), 1079–1097.
Thus most known positively curved manifolds do not admit metrics with positive curvature operator. The simplest example is $CP^n$, $n>1$.
I do not know know much about how sectional curvature pinching can imply positive curvature operator but for example, you could trace references from J.-P. Bourguignon, H. Karcher, Curvature operators: pinching estimates and geometric examples in Ann. Sci. École Norm. Sup. (4) 11 (1978), no. 1, 71–92 where some estimates on the eigenvalues of the curvature operator in terms of sectional curvature pinching can be found.
Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.
Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.
Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details: https://arxiv.org/abs/1403.2117, https://arxiv.org/abs/1412.0039, https://arxiv.org/abs/1511.07899. For example, $(\mathbb C P^n,g_{FS})$ has strongly positive curvature, since $R+\varepsilon (\omega_{FS}\wedge\omega_{FS})>0$ for small $\varepsilon>0$, where $\omega_{FS}$ is the Fubini-Study $2$-form.
The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).