We know that the time dilation formula is given by

$$
begin{align}
triangle t=frac{triangle t_0}{sqrt{1-frac{v^2}{c^2}}}
end{align}
$$

where $triangle t$ is the time measured in the laboratory frame of reference, $triangle t_0$ is the proper time, $v$ is the speed of the moving observer and $c$ is the speed of light in vacuum.

The speed of the spaceship $v$ is

$$
begin{align*}
v=0.75c
end{align*}
$$

The time measured in the frame of reference of the spaceship $triangle t_0$ is

$$
begin{align*}
triangle t_0=1 text{s}
end{align*}
$$

Calculation is straightforward. We simply plug in the given numerical values for the speed of the spaceship $v$ and the time measured in spaceship’s frame of reference $triangle t_0$ into Eq. (1). We get

$$
begin{align*}
boxed{triangle t}&=frac{triangle t_0}{sqrt{1-frac{v^2}{c^2}}}\
&=frac{1s}{sqrt{1-frac{(0.75c)^2}{c^2}}}\
&=boxed{1.51 text{s}}
end{align*}
$$

The time dilation equation is as follows:

$$
begin{gather}
Delta t = dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}}
end{gather}
$$

$textbf{Given: }$

$Delta t_o = 1.00$ s

$Delta t = 1.25$ s

$textbf{Find:}$

speed of the clock (v)

Using Eq (1), we can find an equation that will help us solve for v, such that:

$$
begin{align*}
Delta t &= dfrac{Delta t_{o}}{sqrt{1 – dfrac{v^2}{c^2}}} \
Delta t sqrt{1 – dfrac{v^2}{c^2}} &= Delta t_o \
Delta t^2 left(1 – dfrac{v^2}{c^2}right) &= Delta t_{0}^{2} \
left(1 – dfrac{v^2}{c^2}right) &= dfrac{Delta t_{0}^{2}}{Delta t^2} \
dfrac{v^2}{c^2} &= 1 – dfrac{Delta t_{0}^{2}}{Delta t^2} \
v^2 &= c^2 times left(1 – dfrac{Delta t_{0}^{2}}{Delta t^2} right) \
v &= c sqrt{1 – dfrac{Delta t_{0}^{2}}{Delta t^2}} \
v &= c sqrt{1 – dfrac{(1.00 text{ s})^{2}}{(1.25 text{ s})^2}} \
v &= 0.6c
end{align*}
$$

Hence, the speed of the clock is $boxed{text{0.6 c}}$

$Delta t = 2.00$ s

$v = 0.650$ c

a) Since the speed of the moving clock is comparable to the speed of light (high speed) relativistic effects will be observed as speed increases, $Delta t$ also increases, we can expect that the corresponding elapsed time on the clock at rest will be $boxed{text{less than 2.00 text{ s}}}$

b) Using the time dilation equation we can compute for the elapsed time of the clock at rest as follows:

$$
begin{align*}
Delta t_o &= Delta t times sqrt{1 – dfrac{v^2}{c^2}} \
Delta t_o &= 2.00 text{ s} times sqrt{1 – dfrac{(0.650 c)^2}{c^2}} \
Delta t_o &= boxed{1.52 text{ s}}
end{align*}
$$

b) $Delta t_o = 1.52 text{ s}$