There is no such thing as "rotational work" or "rotational energy". Work is work and energy is energy. In fact, we can go further - "work" and "energy" are just two names for the same thing.

The horizontal force $F$ moves through a horizontal distance $l \sin \theta$ so the work done by the force $F$ is $Fl \sin \theta$. This is not "rotational work" or "translational work" - it is just work.

The centre of mass of the rod is raised by a distance $\frac l 2 (1-\sin \theta)$ so the work done against gravity is $ \frac {mgl} 2 (1-\sin \theta)$. Once again, this is not "rotational work" or "translational work" - it is just work. In fact, you could put this term on the other side of the equation and call it "potential energy gained by the rod" instead of "work done against gravity" - it comes out to the same thing.

The kinetic energy gained by the rod is $\frac {ml^2 \omega^2} 6$. You can think of this as being $\frac 1 2 I_1 \omega^2$ where $I_1$ is the rod's moment of inertia about the stationary pivot. Or you can think of it as being $\frac 1 2 mv^2 + \frac 1 2 I_2 \omega^2$ where $v$ is the velocity of the rod's centre of mass and $I_2$ is the moment of inertia about its centre of mass. You get the same answer either way, because energy is energy.

There is no such thing as "rotational work" or "translational work". Work is work and energy is energy. In fact, we can go further - "work" and "energy" are just two names for the same thing.

The horizontal force $F$ moves through a horizontal distance $l \sin \theta$ so the work done by the force $F$ is $Fl \sin \theta$. This is not "rotational work" or "translational work" - it is just work.

The centre of mass of the rod is raised by a distance $\frac l 2 (1-\sin \theta)$ so the work done against gravity is $ \frac {mgl} 2 (1-\sin \theta)$. Once again, this is not "rotational work" or "translational work" - it is just work. In fact, you could put this term on the other side of the equation and call it "potential energy gained by the rod" instead of "work done against gravity" - it comes out to the same thing.

The kinetic energy gained by the rod is $\frac {ml^2 \omega^2} 6$. You can think of this as being $\frac 1 2 I_1 \omega^2$ where $I_1$ is the rod's moment of inertia about the stationary pivot. Or you can think of it as being $\frac 1 2 mv^2 + \frac 1 2 I_2 \omega^2$ where $v$ is the velocity of the rod's centre of mass and $I_2$ is the moment of inertia about its centre of mass. You get the same answer either way, because energy is energy.