Answer
a) See the figures below.
b) $T_1\gt T_2\gt F_{G,Trout }\gt F_{G,carp}$
Work Step by Step
a) We drew the force diagrams of the two fishes, as you see below.
b) To rank the magnitude of the forces, we need to find the magnitude of each force of them.
We assume that the wire is unstretchable, so both fishes are moving at the same speed.
The net force exerted on the Carp fish is given by
$$\sum F_{y,Carp}=T_1-m_{carp}g-T_2=m_{carp}a_{y,Carp}$$
Thus,
$$a_{y,Carp}=\dfrac{T_1-m_{carp}g-T_2}{m_{carp}}\tag 1$$
And the net force exerted on the trout fish is given by
$$\sum F_{y,Trout }=T_2-m_{Trout }g=m_{Trout }a_{y,Trout }$$
Thus,
$$a_{y,Trout }=\dfrac{T_2-m_{Trout }g}{m_{Trout }}\tag 2$$
Since both of them are having the same acceleration, from (1) and (2), we got the following.
$$\dfrac{T_1-m_{carp}g-T_2}{m_{carp}}=\dfrac{T_2-m_{Trout }g}{m_{Trout }}$$
Plugging the known;
$$\dfrac{60-(1.5\cdot 9.8)-T_2}{1.5}=\dfrac{T_2-(3\cdot 9.8)}{3}$$
Solving for $T_2$;
$$T_2=40\;\rm N$$
Now we know that
$$T_1=\color{blue}{\bf 60}\;\rm N$$
$$T_2=\color{blue}{\bf 40}\;\rm N$$
$$F_{G,carp}=1.5\cdot 9.8=\color{blue}{\bf 14.7}\;\rm N$$
$$F_{G,Trout }=3\cdot 9.8=\color{blue}{\bf 29.4}\;\rm N$$
Therefore,
$$\boxed{T_1\gt T_2\gt F_{G,Trout }\gt F_{G,carp}}$$
