To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables for the quantities of pants and shorts to be made, expressing the objective function in terms of these variables, and formulating the constraints based on the availability of each team.

Let's denote:
- \(x_1\) as the number of pants to be made,
- \(x_2\) as the number of shorts to be made.

The objective function aims to maximize profit. Given that a pant brings in $5 of profit and a short brings in $7 of profit, the objective function can be written as:
\[ \text{Maximize: } 5x_1 + 7x_2 \]

Now, let's formulate the constraints based on the time availability for each operation:
1. Measuring team constraint: \(0.5x_1 + 0.1x_2 \leq 60\)
2. Cutting team constraint: \(0.2x_1 + 0.5x_2 \leq 80\)
3. Sewing team constraint: \(0.7x_1 + 0.6x_2 \leq 75\)

Additionally, the number of pants and shorts cannot be negative:
- \(x_1 \geq 0\)
- \(x_2 \geq 0\)

The symbolic representation in JSON format is:

```json
{
    'sym_variables': [('x1', 'number of pants'), ('x2', 'number of shorts')],
    'objective_function': '5*x1 + 7*x2',
    'constraints': [
        '0.5*x1 + 0.1*x2 <= 60',
        '0.2*x1 + 0.5*x2 <= 80',
        '0.7*x1 + 0.6*x2 <= 75',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

To solve this optimization problem using Gurobi in Python, we'll use the following code:

```python
from gurobipy import *

# Create a new model
m = Model("Tailor_Shop_Optimization")

# Define variables
x1 = m.addVar(name="pants", vtype=GRB.CONTINUOUS, lb=0)
x2 = m.addVar(name="shorts", vtype=GRB.CONTINUOUS, lb=0)

# Set the objective function
m.setObjective(5*x1 + 7*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(0.5*x1 + 0.1*x2 <= 60, name="measuring_team")
m.addConstr(0.2*x1 + 0.5*x2 <= 80, name="cutting_team")
m.addConstr(0.7*x1 + 0.6*x2 <= 75, name="sewing_team")

# Optimize the model
m.optimize()

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