To solve the given optimization problem, we first need to define the variables and the objective function symbolically. Let's denote:
- $x_1$ as the amount of the first mix (containing 30% honey-roasted almonds and 70% chocolate-covered almonds) to be prepared in kg.
- $x_2$ as the amount of the second mix (containing 40% honey-roasted almonds and 60% chocolate-covered almonds) to be prepared in kg.

The objective function, which is to maximize profit, can be represented as:
\[ \text{Maximize:} \quad 12x_1 + 15x_2 \]

Given the constraints:
- The store has 100 kg of honey-roasted almonds. Since $x_1$ contains 30% honey-roasted almonds and $x_2$ contains 40%, we have: 
\[ 0.3x_1 + 0.4x_2 \leq 100 \]
- The store has 150 kg of chocolate-covered almonds. Since $x_1$ contains 70% chocolate-covered almonds and $x_2$ contains 60%, we have:
\[ 0.7x_1 + 0.6x_2 \leq 150 \]
- Non-negativity constraints for both mixes, since we cannot prepare a negative amount of either mix:
\[ x_1 \geq 0, x_2 \geq 0 \]

Thus, the symbolic representation of the problem in JSON format is:

```json
{
    'sym_variables': [('x1', 'amount of first mix in kg'), ('x2', 'amount of second mix in kg')],
    'objective_function': '12*x1 + 15*x2',
    'constraints': [
        '0.3*x1 + 0.4*x2 <= 100',
        '0.7*x1 + 0.6*x2 <= 150',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

Now, let's implement this problem using Gurobi in Python:

```python
from gurobipy import *

# Create a model
m = Model("Honey_and_Chocolate_Almonds")

# Define the variables
x1 = m.addVar(name="x1", lb=0)
x2 = m.addVar(name="x2", lb=0)

# Set the objective function
m.setObjective(12*x1 + 15*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(0.3*x1 + 0.4*x2 <= 100, name="honey_roasted_almonds")
m.addConstr(0.7*x1 + 0.6*x2 <= 150, name="chocolate_covered_almonds")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"x1 = {x1.x}")
    print(f"x2 = {x2.x}")
    print(f"Maximum Profit: {m.objVal}")
else:
    print("No optimal solution found")
```