Is it possible to solve for the tension in rope and the acceleration of the 2 blocks without knowing the friction ?
No.
Without knowing the value of friction, it is impossible to calculate the acceleration of the $1\ kg$ block, because the acceleration itself depends on the friction.
Let us call the $1\ kg$ block as block 1 and $2\ kg$ block as block 2.
There is a systematic way to solve your problem, and it is by considering every part of the system one by one.
Consider the following points-
1. Either the $1\ kg$ block will move towards right, or it won't move at all.
2. Let us calculate the net force on the block of mass $2\ kg$. Let the system's acceleration be $a$.
$$2g\ \text{sin}\ 53^{\circ}+16-T_1=2a\ ...(1)$$
which gives us-
$$T_1=32 - 2a\ ...(2)$$ considering $g=10\ m/s^2$.
From this equation, we see that if $a=0$ i.e. for the system at rest, the tension $T_1$ has a maximum value which is $32\ N$.
3. The pulley either rotates clockwise, or does not rotate at all. ( Again, because of point 1). Therefore, the torque towards the clockwise direction is always greater than or equal to the torque towards anticlockwise direction. This gives us- $T_1 \geq T_2.$
Now, let us suppose that block 2 is at rest, i.e. $a=0$ in $(2)$. This gives us $T_1=32\ N$. Now, because block 2 is at rest, block 1 is also at rest, due to the string's length being conserved. This tells us that the pulley is not rotating, which means-
$$T_2=T_1=32\ N$$.
But the limiting value of friction on block 1 is $2\ N$. Hence, it is not under equilibrium.
We assumed that block 2 does not move, and it turned out that in that case, block 1 will move. We thus have a contradiction. But where does this contradiction stem from?
The assumption that block 2 can stay at rest.
We thus conclude that block 2 can not stay at rest, which means block 1 will move too, giving rise to limiting value of friction i.e. $2\ N$.