Answer
(a) less than
(b) $\pi \sqrt{\frac{l}{g}}+\pi \sqrt{\frac{L}{g}}$
(c) $1.5s$
Work Step by Step
(a) We know that the time period decreases without the peg because the length of the pendulum will be shortened and the time period is directly proportional to the length.
(b) The required time period is the sum of the half period of the shortened pendulum and the period of the long pendulum -- that is, $T=\pi \sqrt{\frac{l}{g}}+\pi \sqrt{\frac{L}{g}}$
(c) We can find the required time period as
$T=\pi \sqrt{\frac{l}{g}}+\pi \sqrt{\frac{L}{g}}$
We plug in the known values to obtain:
$T=\pi \sqrt{\frac{0.25m}{9.81m/s^2}}+\pi \sqrt{\frac{1.0m}{9.8m/s^2}}$
$T=1.5s$
