For circular orbits, we get to use this short derivation:
The centripetal acceleration holding the satellite on its orbit is $ \ a_c = \frac{v^2}{R} \ . $ It is the force of Earth's gravity at that distance from the center of the planet that provides that acceleration, $ \ a_g = \frac{GM}{R^2} \ . $ Here, $ \ R \ $ is the radius of the orbit, $ \ M \ $ is the mass of Earth, and $ \ G \ $ is the Universal Gravitational Constant...constant...consta--
We may thus set these two accelerations equal and solve for the "circular orbital velocity", $ \ v_c = \sqrt{\frac{GM}{R}} \ . $ You will need $ \ G = 6.67 \cdot 10^{-11} \ $ in System Internationale (SI) units. Don't forget that the orbital radius $ \ R \ $ is Earth's radius plus the orbital altitude. The distances must be in meters, as given; the mass in kilograms is fine. This answers question 2 (in $ \ \frac{\text{m.}}{\text{sec.}^2} \ $ ) , and question 1 is the circumference of the orbit divided by the circular orbital speed.
I will note, incidentally, that the orbital altitude given in the problem does not agree with that given in the articles on GPS ($ \ 2.02 \cdot 10^7 \ $ m.), so you will not get the period given there of half a sidereal day, or 43,082 seconds. [The altitude used is peculiar to that "constellation" of satellites, so this is often referred to as a "GPS orbit". Geosynchronous satellites complete their orbits in a full sidereal day, so they are at the still higher altitude of $ \ 3.58 \cdot 10^7 \ $ m.]
You would need to consider two cases, calculating a formula for the speed of an object at its aphelion and perihelion.
You could show that the position of an object in orbit satisfies
$$r(\theta) = \frac{L^{2}}{GMm^{2}(1+e\cos \theta)}$$
Where $L$ is the angular momentum of the object and $e$ is the eccentricity of the orbit (for a circular orbit, $e=1$).
Let me sketch out the aphelion condition. Namely the point at which $\theta = 0$, also $r=a(1-e)$. Viz,
$$a(1-e)=\frac{L^{2}}{GMm^{2}(1+e)}$$
From which we obtain
$$\left(\frac{L}{m}\right)^{2}=GMa(1-e^{2})$$
I quote without proof the formula for the area swept out by an object in terms of its angular momentum $L$ and period $P$.
$$\frac{\pi a b}{P}=\frac{L}{2m}$$
Substituting in for $L^{2}/m^{2}$
$$\frac{\pi^{2}a^{2}a^{2}(1-e^{2})}{P^{2}}=\frac{a(1-e^{2})GM}{4}$$
Simple rearrangement gives a formula for $P^{2}$ in terms of $a^{4}$. Now you must repeat at the perihelion, taking note of what the position $r$ of the object would be at that point.
Best of luck,
Bacon.
Best Answer
$T={1\over f}=1 year=365day*24{h\over day}*60{min\over h}*60{s\over min}=31536000s\approx3.15*10^7s$