Q1: How fast are S and P moving (in Q's frame)?
Any two observers will always agree on their relative velocity. So if $S$ sees $Q$ moving at $c/2$ then $Q$ sees $S$ moving at $-c/2$.
Calculating the velocity of $P$ in $Q$'s frame is harder. You could just use the formula for relativistic addition of velocities to get:
$$ v'_p = \frac{v_p - v_q}{1 - \frac{v_pv_q}{c^2}} \tag{1} $$
but this brings no understanding with it, so let's use the Lorentz transformations:
$$\begin{align}
t' &= \gamma \left( t - \frac{vx}{c^2}\right) \\
x' &= \gamma \left( x- vt \right)
\end{align}$$
Start in $S$'s frame. In this frame $P$ is moving with velocity $v_p$ so in a time $t$ $P$ moves from the point $(0, 0)$ to $(t, v_pt)$. Well use the usual convention that the origins of all frames coincide at time $t = 0$, so the point $(0, 0)$ is the same in all frames. We just have to calculate where the point $(t, v_pt)$ is in $Q$'s frame. The velocity of $Q$ is $v_q$, so just plugging the values into the Lorentz transformations we get:
$$\begin{align}
t' &= \gamma \left( t - \frac{v_qv_pt}{c^2}\right) \\
x' &= \gamma \left( v_pt - v_qt \right)
\end{align}$$
The velocity of $P$ in $Q$'s frame is simply $x'/t'$ so:
$$\begin{align}
v'_p &= \frac{\gamma(v_pt - v_qt)}{\gamma(t - \frac{v_qv_pt}{c^2})} \\
&= \frac{v_p - v_q}{1 - \frac{v_qv_p}{c^2}}
\end{align}$$
which is the same as equation (1).
Q2: Does Q see S's clocks as moving slower or faster?
Every observer sees the clocks of a moving observer running slow. So $S$ sees $Q$'s clocks running slow by a factor of $\gamma$, but $Q$ sees $S$'s clocks running slow by the same factor of $\gamma$. This is the origin of the famous twin paradox.
Q3: Does Q see P's clocks as moving at $\gamma_p$ or at $\gamma_q/\gamma_p$?
Neither.
If you take equation (1) and feed in $v_p = 0.25c$ and $v_q = 0.5c$ you get the velocity of $P$ in $Q$'s frame as:
$$ v'_p \approx 0.286 c $$
If you work out the corresponding value of $\gamma$ you get:
$$ \gamma'_p \approx 0.958 $$
Q4: How fast are S and Q moving?
In Q1 we worked out that $Q$ sees $P$ moving at about $0.286c$, so $P$ sees $Q$ moving at $0.286c$.
Likewise $P$ sees $S$ moving at $0.25c$.
Q5: Does P see a difference in S and Q's clocks?
Yes, because in $P$'s frame $S$ and $Q$ are moving at different speeds.
Once S, P and Q are reunited. What is the order of the clocks? Which one is earliest, which one is latest and which one is in between?
To do the calculation would be hard because you'd have to calculate the time dilation in the acceleration phases. Fortunately if you just want the relative order that's easy because you can get it from the following thought experiment:
We send $P$ and $Q$ off, and they both return a time $t$ later. We know that time dilation means their elapsed times, $t_p$ and $t_q$, will both be less than $t$. Suppose we make the velocity of $P$ very small, then $t_p \approx t$, and if we take the limit of $P$ not moving at all we'd have $t_p \approx t$. As we increase the velocity of $P$ the time $t_p$ will start falling smoothly with increasing velocity. The exact form of the function $t_p(v_p)$ will be complicated, but we expect it to fall smoothly towards zero as $v_p \rightarrow c$.
So because $v_q > v_p$ we know $t_q < t_p$, so $S$'s clock shows the most elapsed time and $Q$'s clock shows the least elapsed time.
