I think we need to assume that the concave lens is placed up against the convex lens, both of which are $65$ cm from the object. Otherwise, there is not enough information given to solve the problem.
Hint:
When lenses are against each other, their strengths (the reciprocals of their focal lengths) are added. Thus,
$$
\frac{1}{65\text{cm}}+\frac{1}{20\text{cm}}=\frac{1}{f_{\text{convex}}}
$$
$$
\frac{1}{65\text{cm}}+\frac{1}{75\text{cm}}=\frac{1}{f_{\text{convex}}}+\frac{1}{f_{\text{concave}}}
$$
Hint : Friction opposes tendency to move. Tension is produced if the string is stretched $very$ slightly. So, increase friction to maximum and then tension will act if necessary.
Ironically, you are thinking absolutely right. Give yourself a cookie.
From part $a$, we know that the blocks will be at rest at all angles below that.
You are also right as at very small angles there is no need of tension and we can ignore it to solve for, again, an angle condition. You have done excellent work. Congrats.
Now we come to the middle angles. Oh... they drive you insane, don't they?
Let's start. We can start our analysis from 2 blocks, 1 will give a contradiction and other will give a result, but I will start with the one giving contradiction. This will help you.
All angles are in degrees :
$\theta=35 $
Lets start by analysing Block A (No racism intended)
Gravity is trying to pull it down : $5*10*\sin(35)N=28.67N$
Friction comes to the rescue(up) : $50*\cos(35)N=16.38N$ // read my hint to know why friction is put max here
As it is at rest, $T=12.29N$
Now Block B is also at rest,
Weight = $114.71N$
max f= $81.92N$
$16.38+114.71=12.29+f$
$f=118.8N$
OOPS, it exceed max value. So, Lets start by analysing Block B. (I love alliteration)
Gravity trying : $114.71N$
Friction comes to the rescue(up) : $81.92N$
You can take from here I guess. calculate tension. Note that you have to revise your calculation for tension again as reaction friction force will be provided by A. Better assume it $f$ from starting FBD of B.
This will yield the correct answer. Friction will be less than max value for upper block. In most cases, You should start analysing with heavier block(my experience). Hope your doubts are cleared.
Best Answer
Actually, the forces $\vec{F_1}, \vec{F_2}, \vec{F_3}$ defined by the points $(3,4,0)$ , $(-3,0,5)$ and $(0,-4,-5)$ are coplanar.
Even though $\vec{F_1}, \vec{F_2}, \vec{F_3}$ algebraically sum to $\vec 0$ they are nevertheless coplanar.
In fact, if you do a little algebra (solve three simultaneous equations) or calculate the cross product $(\vec{F_2}-\vec{F_1})\times (\vec{F_3}-\vec{F_1})$ (which gives you the vector normal to this plane) you deduce that $(3,4,0)$, $(-3,0,5)$ and $(0,-4,-5)$ lie on the plane defined by the equation $$20x-15y+12z=0$$
So the rule still holds and you would actually need $3+1$ non coplanar forces for equilibrium, where the forth force would need to be equal and opposite to the sum of the first three forces.
Note that three coplanar forces, or even two coplanar forces can sum to zero.