I is true. Let $r\text{ cm}$ be the radius of the sector $OAB$.
$\begin{array}{rcl}
\pi r^2 \times \dfrac{90^\circ}{360^\circ} & = & \pi \\
r^2 & = & 4 \\
r & = & 2
\end{array}$
$\therefore$ the radius of the sector $OAB$ is $2\text{ cm}$.
II in not true. The perimeter of the sector $OAB$
$\begin{array}{cl}
= & 2(2) + 2\pi(2) \times \dfrac{90^\circ}{360^\circ} \\
= & (4 + \pi) \text{ cm}
\end{array}$
III is true. Since $\angle AOB =90^\circ$, then $AB$ is a diameter of the circle passing through $O$, $A$ and $B$.
The radius of the circle
$\begin{array}{cl}
= & \dfrac{1}{2} AB \\
= & \dfrac{1}{2} \sqrt{2^2+2^2} \\
= & \dfrac{\sqrt{8}}{2} \text{ cm}
\end{array}$
Hence, the area of the circle
$\begin{array}{cl}
= & \pi \left(\dfrac{\sqrt{8}}{2}\right)^2 \\
= & 2\pi\text{ cm}^2
\end{array}$
