Finding tension in pulley system

I'm struggling with part C of this question: https://isaacphysics.org/questions/ft_mech_stage1_q4_2018?board=y12ms2021_22_week15&stage=a_level

(acceleration of particle $R = 4.21$. Worked out in part B)

I tried drawing free-body diagrams for particles P, Q and R. I've also tried writing $F=ma$ for each particle.

I ended up with these 3 equations:

$T-mg=ma$ (P)

$0.45g-2.52=0.45a$ (R)

$0.05g-T=0.05a$ (Q)

I don't know where to go from here - I tried assuming that R and Q would have the same acceleration, and used the equations to work out tension $= 0.28~\text{N}$ (this was wrong).

It would be really helpful if someone could point me in the right direction or tell me where I'm going wrong. Thanks

(I'm doing A Level maths and physics if that helps)

Firstly, I got $a=4.2,\ $ not $a=4.21:$

$(b)\quad F=ma$ on particle $R\ $: $0.45 \times 9.8 - 2.52 = 0.45a,\ $ so $\ a = \frac{0.45 \times 9.8 - 2.52}{0.45} = 4.2\ $ ms$^{-2}.$

Secondly, your equation for $Q$ is not correct because particle $R$ is pulling particle $Q$ downward with a force of $2.52$N. You have left this out of your equation when you shouldn't have...

Finally, all you then have to do is to substitute $a=4.2\ $ into your equation for $Q$ and rearrange to get the value of $T.$

$(c)\quad F=ma\ $ on particle $Q:\ 2.52 + 0.05\times 9.8 - T = 0.05 \times 4.2 \implies T = 2.8$N.