>>1147195411.154 kg
You don't only need the friction force from block 2, you also need to add the gravitational force component acting on the block at the angle of 35.5 degrees. We'll call that combined friction and weight force F(p).
F(p) = [F(g) * Cos(35.5)] + uF(N)
F(p) = [12.1 * 9.8 * Cos(35.5)] + [0.2 * 12.1 * 9.8 * Sin(35.5)]
F(p) = 95.538 + 13.772
F(p) = 109.31 N
Since block 2 is moving at a constant speed, we know the weight force of block 1 must equal F(p) of block 2. So by rearranging F=ma, we can find the mass of block 1.
m = F / a
m = 109.31 / 9.8
m = 11.154 kg
Think about it. Block 2 is on a slope, that means a good chunk of its weight force is being counterbalanced by the normal force. The surface doesn't have enough friction to make up for this reduction in effective weight force. Taking this idea to the extreme, imagine a 1000 kg mass on a zero friction surface at some angle, with a 1 kg block hanging from the other end of the pulley. There exists an angle at which the weight force of the 1kg block will exceed the effective weight force of the one tonne block, and pull it up the incline. This question is literally all about mechanical advantage and if you can't figure that out I suggest you seriously think about how forces work and balance to try to understand more deeply. Your objection that block 1 should weight more than block 2 is about as intelligent as saying a kilogram of feathers weigh less than a kilogram of gravel.
