The same metre rule, 1.00 m long, can hang from a hole drilled anywhere along it…

(a) About the middle, $I_c = \tfrac{1}{12}mL^2$. The parallel axis rule gives $I = \tfrac{1}{12}mL^2 + md^2$ about the hole. Put it in $T = 2\pi\sqrt{I/mgd}$, and $m$ cancels.

(b) $T$ is smallest when $\dfrac{L^2/12}{d} + d$ is smallest. That happens when the two parts are equal: $d^2 = \dfrac{L^2}{12}$, so $d = \dfrac{L}{\sqrt{12}} = \dfrac{1.00\ \text{m}}{3.46} = 0.289\ \text{m}$.

(c) Then $\dfrac{L^2/12 + d^2}{d} = 2d = 0.577\ \text{m}$, and $T = 2\pi\sqrt{\dfrac{0.577\ \text{m}}{9.8\ \text{m/s}^2}} = 1.53\ \text{s}$.