T is equal to 5gTcos 53 for left side and 5gTcos37 for the right side
As posed, you have three unknowns - $a1$, $a2$, and $a3$. The tension on the cable is everywhere uniform, but the accelerations are different because the static force components affecting the tension are different.
$F=ma$ at each mass. So for mass $m1$, the tension is
$m1*(g*cos53 +a1)$. Likewise
$m2*(g*cos37 + a2)$ and
These are all equal. And the length of the line doesn't change.
You can assume that you start from a stationary condition (all velocities are initially zero) - that doesn't change the accelerations. You can choose any initial length that like - that doesn't change the accelerations either.
So $m1(g*cos53 + a1) = 1/2 m3(g*cos90 + a3) = m2(g*cos37 +a2) $ from equal tension.
And $a1 = 2*a3 - a2$ from constant length.
$ 5 kg (0.600g + a1) = 3 kg (1g + a3) = 5 kg ( 0.799g + a2)$
$0.01 kg + 5 kg * a1/g = 3 kg * a3/g = 0.993 kg + 5 kg * a2/g$
$0.0033 + 1.666 a1/g = a3/g = 0.331 + 1.666 a2/g$
$0.0033 + 1.666 (2a3 - a2)/g = a3/g$
$0.0033 - 1.666 a2/g = -2.332 a3/g$
$-0.0014 + 0.7144 a2/g = a3/g $
$-0.0014 + .7144 a2/g = 0.331 + 1.666 a2/g$
$-0.3324 = .9516 a2/g$
$a2 = -0.349 g$
$a3 = -0.2509 g$
$a1 = -0.1529 g$