Relative motion and time When someone reaches to a speed which is close to the speed of light with respect to earth, will he see the things actually moving faster than when he is in low speeds?
 A: I'm of two minds about answering this because, on the one hand, it's a straightforward homework exercise, but on the other hand, it seems to me that the other two posted answers are at best misleading.
Michele Grosso's answer says that earth clocks tick slowly in the spaceship's frame, but that's not what I think you're asking.  I interpret your question to mean this:  "If a clock on earth ticks once per second in its own frame, will the light from those ticks reach the spaceship more or less frequently than once per second?".  
I'll adopt Árpád Szendrei's setup:  The earth is at $(x=0,y-0)$ and the spaceship is at $(1,1)$ moving along the line $y=1$ with speed, say, $1/2$.  
If the ship moves rightward, the answer is clear:  First, the earth clock ticks slowly in the spaceship's frame, and second, light from each successive tick takes longer to reach the spaceship.  So the traveler certainly sees the earth clock tick in slow motion.
If the ship moves leftward, the two effects go in opposite directions, so one needs a bit of calculation.  
First, calculate  everything in the earth frame:  
The earth clock ticks at time $0$ and light from this click happens to hit the traveler just as he passes the point $(1,1)$.  This light has traveled a distance $\sqrt2$ so the traveler sees it at $(x=1,y=1,t=\sqrt2)$.  
Now the earth clock ticks at time $1$ and light from this tick reaches the traveler at time $T+1$.  Thus the interval between the two arrivals is $T+1-\sqrt2$, and the traveler has now reached location $(X,1)$ where $X=1-\left({T+1-\sqrt2\over2}\right)$.  Thus the light has traveled a distance $\sqrt{X^2+1}$ in time $T$, giving $\sqrt{X^2+1}=T$, whence $T\approx 1.176$ and $X\approx .619$.  Thus the arrival occurs at $(x=.619,y=1,t=2.176)$.
Now Lorentz-transform the two arrivals into the ships frame.  The arrival of the first tick at ($x=1,y=1,t=\sqrt2$) transforms to $t'\approx 2.2$ and the arrival of the second tick transforms to $t'\approx 2.87$.  Thus the time between the arrivals is $2.87-2.2<1$, which means the spaceship sees the clock speeded up.
I chose to do this with explicit numbers rather than generic locations and velocities, in hopes that it would be more readable.  God knows I might have screwed up the arithmetic, but something very like this should be right.
A: To your question, let us suppose there is a clock on earth, and another clock with the one traveling.
Let just use SR for this case and we disregard the effects of GR.
As per the comment, it is very important to see that it matters whether the traveler is closing up to earth or moving away from it. Even if he is moving perpendicular compared to earth (so horizontally seen from earth), it will matter whether he is closing up or receding.
In case he is moving closer to earth (at near the speed of light), then he will see clocks tick faster on earth.
But in case he is moving away from earth, then time dilation might not affect his view of the clock on earth, so he might not see the clock on earth tick faster. This is because it will take more time for light to reach him when he is receding.
But now, for your question, let's disregard this effect, and assume that he is not closing up nor receding from earth.
Let's assume that he is moving perpendicular to earth (so neither moving directly towards nor away from earth), so that it is seen as a horizontal move as seen from earth.
You are asking if the person traveling would see the clock on Earth tick slower or faster whether he is traveling slow or fast.
The person traveling near the speed of light will see the clock on earth tick faster then his own clock with him.
The same person traveling slow (at non-relativistic speed) will see no real difference between the clock on earth and his own clock.
So the answer to your question whether he will see things happen faster on earth if he moves near the speed of light is yes.
He will see things happen faster on earth, and the clock on earth tick faster if he moves near the speed of light, compared when he moves slower.
So yes, as per SR, the person's speed, especially when he moves near the speed of light, will make things on earth seem to happen faster (clock tick faster) compared to when he is moving slow.
A: In SR (special relativity) time is running at a different pace if measured in inertial reference frames in relative motion. A stationary observer (observer in its own rest frame) measures a time $t$ running faster than the proper time $\tau$ of a moving frame. The well known relation is:
$\Delta t = \gamma \Delta \tau$ time dilation
where:
$\gamma = 1 / \sqrt{1 - v^2/c^2}$ Lorentz factor  
As the relative velocity $v$ approaches $c$ the Lorentz factor diverges to infinity and the stationary observer measures the clock of the moving frame to tick slower and slower. Of course, the description is symmetric if measured by the moving frame.  
To answer to the question, a spacecraft moving at a relativistic speed vs. earth would measure events on the earth occurring slower. That disregarding whether it is going closer or farther from the earth, as the Lorentz factor depends only on the norm of the relative velocity.  
Note: A different issue is the relativistic Doppler effect. In that case going closer would mean a blueshift of the radiation, while going farther a redshift vs. the radiation frequency as measured in the earth reference frame.
A: Albert Einstein in his celebrated paper “On the Electrodynamics of moving bodies” teaches us, that moving observer (I emphasize – he speaks about moving observer) will see, that clock “at rest” is ticking $\gamma$ times faster than his own.
Chapter 7, Theory of Doppler principle and aberration (p. 16)
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
Let’s assume that a clock on Earth emits green monochromatic light (spherical waves) of proper frequency $f_s$. Frequency $f_s$ is nothing but speed of ticking. 
Einstein gives the following formula for relativistic Doppler shift:
$$ f_0 = \gamma (1 – \frac {v \cos \theta_s} {c}) f_s \quad (1)$$
This formula shows, that $f_s$ is gamma times lower, than $f_o$, i.e. from the point of view of moving observer, clock at rest is ticking faster. It is obvious, since clock of the observer (or the observer himself) turn into dawdler, everything around him appears running as in fast forward mode.
However, relativistic Doppler shift also includes contribution due to motion of observer.  This contribution disappears, when observer and the source are at points of closest approach.
At the moment of closest approach (right above Earth) observer neither approaches nor recedes from the source, and "sees" or better to say "measures" only contribution of time dilation into the Doppler shift.
At this instant, observer moves tangentially to the wave front and angle  $\theta_s = \pi/2$. Accordingly, $\cos \theta_s = 0$. In classic case there should not be any Doppler shift, but due to dilation of observer’s clock this frequency shift is equal to:
$$f_0= \gamma f_s$$
That means, moving with velocity very close to that of light observer measures, 
that radiation turns very violet, or clock of the source is ticking very, very fast.
This effect (measurement of frequency at points of closest approach) is also known as Transverse Doppler Effect. 
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
