$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}$Let $F$ be velocity vector field of fluid on $\Reals^3$ defined by $F(x,y,z) = -y\Vec{i} + x\Vec{j}$.
(A) Show that $F$ is rotational.
(B) Show that if we place a cork in this field it will resolve in a plane parallel to the $(x, y)$ plane in a circular trajectory.
(C) In what direction does the cork revolve?
I know that that $F$ is rotational means $\nabla \times F \neq 0$ then how we going solve (B) and (C).
I don't know what a "rotational field" is. In the first place your $F=(-y,x,0)$ is a plane field: It's $z$-component is $\equiv0$, and the two other components do not depend on $z$. Therefore we understand everything about $F$ if we understand what happens in the $(x,y)$-plane.
It is a basic fact of plane analytic geometry that $$j:\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad(x,y)\mapsto(-y,x)$$ is a $90^\circ$ degree rotation counterclockwise. This implies that the field vector $F(x,y)=(-y,x)$ is obtained from the "position vector" $(x,y)$ by a $90^\circ$ rotation. In other words: $F(x,y)$ is tangent to the concentric circle through $(x,y)$. From this we can intuitively deduce that your cork swims in a circle counterclockwise. But there is also a formal proof:
Let $t\mapsto\bigl(x(t),y(t)\bigr)$ be the orbit of the cork, and consider the auxiliary function $$\Phi(t):=x^2(t)+y^2(t)\ .$$ Since $(\dot x,\dot y)=F=(-y,x)$ one has $$\Phi'(t)=2x(t)\dot x(t)+2y(t)\dot y(t)=-2x(t)y(t)+2y(t)x(t)\equiv0\ .$$