Answer
a) $\rm 24\;h$
b) $0.223\;\rm m/s^2$
c) $0\;\rm N$
Work Step by Step
a) The period of a geosynchronous satellite is 24 hours since the Earth completes one full rotation around its axis every 24 hours.
$$T=\color{red}{\bf 24}\;\rm h$$
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b) The free-fall acceleration at this point is the radial acceleration of the satellite which is given by
$$a_r=g=\dfrac{v^2}{R_o}\tag 1$$
So, we know that the velocity of this satellite is given by
$$v=\dfrac{2\pi R_o}{T}$$
Plugging into (1);
$$a_r=g=\dfrac{4\pi^2 R_o^2}{T^2R_o} $$
$$ g=\dfrac{4\pi^2 R_o }{T^2 } $$
where $R_o=R_E+h$ whereas $R_E$ is earth's radius while $h$ is the satellite height.
$$ g=\dfrac{4\pi^2 (R_E+h)}{T^2 } $$
Plugging the known;
$$ g=\dfrac{4\pi^2 ([6.37\times 10^6]+[3.58\times 10^7])}{(24\times 60^2)^2} $$
$$g=\color{red}{\bf 0.223}\;\rm m/s^2$$
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c) Since the satellite is at a constant height and orbiting the Earth at a constant speed, it is under an endless free-fall process which means that its weight is zero.
Also, there is no normal force exerted on the satellite, so its wright is zero.
$$W=\color{red}{\bf 0}\;\rm N$$
