Let's solve the given problem step by step.

1. **Equation to Solve for \(x\):**

\[
\sin x - \cos x = \frac{1}{2}
\]

2. **Expressions to Calculate:**

- \(A = \sqrt{\sin^4 x + \cos^4 x}\)
- \(B = \sqrt{\sin^6 x + \cos^6 x}\)
- \(C = \sqrt{\tan^2 x + \cot^2 x}\)

### Step-by-Step Solution:

#### Step 1: Solve for \(x\)

The equation \(\sin x - \cos x = \frac{1}{2}\) doesn't immediately provide a simple solution. We might use the identity \(\sin x = \sqrt{\frac{1 + \cos 2x}{2}}\) and \(\cos x = \sqrt{\frac{1 - \cos 2x}{2}}\) to transform and eventually solve the equation. However, let's assume the values that satisfy the original equation.

#### Step 2: Calculate \(A\)

\[
\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x
\]
Using \(\sin^2 x + \cos^2 x = 1\), then:
\[
A = \sqrt{1 - 2\sin^2 x \cos^2 x}
\]
Using \(\sin x \cos x = \frac{\sin 2x}{2}\):
\[
A = \sqrt{1 - \frac{1}{2}\sin^2 2x}
\]

#### Step 3: Calculate \(B\)

\[
\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)
\]
\[
B = \sqrt{1 - \frac{3}{4}\sin^2 2x}
\]

#### Step 4: Calculate \(C\)

\[
\tan^2 x + \cot^2 x = \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\sin^2 x}
\]
\[
C = \sqrt{\frac{\sin^4 x + \cos^4 x}{\sin^2 x \cos^2 x}}
\]

#### Conclusion:

Since exact values of \(x\) are not given, completing numerical calculations for \(A\), \(B\), and \(C\) requires either knowledge of \(\sin x\) and \(\cos x\) that meet the initial condition or specifying an approximate solution for clarifying results.