In (1) a question is asked about averaging a certain allocation.
Let \(i\) be the set of prices to allocate to buckets \(j\). As each price can go to at most one bucket, it makes sense to use a binary variable:
\[x_{i,j} = \begin{cases} 1 & \text{if price $i$ is allocated to bucket $j$}\\ 0 & \text{otherwise}\end{cases}\] 
The objective is to minimize the spread of the average prices seen in each bucket. A (nonlinear) model can look like:
\[ \bbox[lightcyan,10px,border:3px solid darkblue] {\begin{align} \min\>& \left(\max_j \mu_j – \min_j \mu_j \right)\\ & \mu_j = \frac{\displaystyle\sum_i p_i x_{i,j}}{\displaystyle\sum_i x_{i,j}}\\ &\sum_j x_{i,j} = 1 & \forall i\\ &x_{i,j} \in \{0,1\} \end{align}}\] 
Here the variable \(\mu_j\) is the average price in bucket \(j\). We have two nonlinearities in this model: the objective, and the calculation of \(\mu_j\).
The objective is easily linearized:
\[\begin{align} \min\> & \mu_{max} – \mu_{min}\\ & \mu_{max} \ge \mu_j & \forall j\\ & \mu_{min} \le \mu_j & \forall j \end{align}\] 
But what about the calculation of the average? We can write:
\[\mu_j \sum_i x_{i,j} = \sum_i p_i x_{i,j} \] 
This does not help much. If we introduce \(z_j = \sum_i x_{i,j}\) then the lefthand side is \(\mu_j z_j\) which is still not easy to linearize. However, a better approach is to write
\[\sum_i \mu_j x_{i,j} = \sum_i p_i x_{i,j} \] 
The expressions \(\mu_j x_{i,j}\) are still nonlinear, but they are a multiplication of a binary variables times a continuous variable. This we know how to linearize (2).
The complete linearized model can look like:
\[\bbox[lightcyan,10px,border:3px solid darkblue]{\begin{align} \min\> & \mu_{max} – \mu_{min}\\ & \mu_{max} \ge \mu_j & \forall j\\ & \mu_{min} \le \mu_j & \forall j \\ & \sum_i y_{i,j} = \sum_i p_i x_{i,j} & \forall j\\ & y_{i,j} \le U x_{i,j}\\ & y_{i,j} \le \mu_j\\ & y_{i,j} \ge \mu_j – U(1x_{i,j})\\ &\sum_j x_{i,j} = 1 & \forall i\\ &y_{i,j} \in [0,U]\\ &x_{i,j} \in \{0,1\} \end{align}}\] 
where \(U=\displaystyle \max_i p_i\) (this is data). The variable \(y_{i,j}\) can be interpreted as \(y_{i,j}=\mu_j x_{i,j}\).
In a real application we would have other restrictions as well. The most obvious is that we want to allocate at least one price to a bucket. This is simply done by requiring \(\displaystyle\sum_i x_{i,j} \ge 1\). Here is some example output while including this constraint:
 53 SET i prices i1 , i2 , i3 , i4 , i5 , i6 , i7 , i8 , i9
j1, j2, j3, j4, j5
i1 1.717, i2 8.433, i3 5.504, i4 3.011, i5 2.922
j1 j2 j3 j4 j5 i1 1
j1 4.401, j2 4.552, j3 3.872, j4 4.007, j5 3.498
j1 j2 j3 j4 j5 i1 4.401

Side note: What would happen if we allow zero prices to go into a bucket? The model would not restrict \(\mu_j\) for such a bucket. In extremis we can put all prices in the first bucket (with average price \(\mu_1\)), and then have all the remaining buckets empty. That would allow the model to pick \(\mu_j=\mu_1\) i.e. no spread. That is exactly what I got when I tried this (by dropping the constraint \(\displaystyle\sum_i x_{i,j} \ge 1\)). The results with the same data as before:
 53 VARIABLE x.L assignment
j1
i1 1
i2 1
i3 1
i4 1
i5 1
i6 1
i7 1
i8 1
i9 1
i10 1
 53 VARIABLE mu.L averagej1 4.156, j2 4.156, j3 4.156, j4 4.156, j5 4.156
 53 VARIABLE y.L average or zeroj1
i1 4.156
i2 4.156
i3 4.156
i4 4.156
i5 4.156
i6 4.156
i7 4.156
i8 4.156
i9 4.156
i10 4.156
 53 VARIABLE mumin.L = 4.156
VARIABLE mumax.L = 4.156
The same trick can be used to linearize the problem stated in (3).
References
 Fair allocations to buckets with constraints, http://stackoverflow.com/questions/43662264/fairallocationtobucketswithconstraints
 Multiplication of a continuous and a binary variable, http://yetanothermathprogrammingconsultant.blogspot.com/2008/05/multiplicationofcontinuousandbinary.html
 Convert nonlinear objective function to linear objective function, http://stackoverflow.com/questions/43591174/convertnonlinearobjectivefunctiontolinearobjectivefunction