a) \( \sin ^{2}\left(150^{\circ}+x\right)-\cos (-x) \cdot \tan \left(90^{\circ}-x\right) \sin \left(540^{\circ}+x\right. \)

Let's first simplify the expression \( \sin ^{2}\left(150^{\circ}+x\right)-\cos (-x) \cdot \tan \left(90^{\circ}-x\right) \sin \left(540^{\circ}+x\right) \). We start by breaking down components: 1. **Sine Function**: \[ \sin(150^{\circ}+x) = \sin(150^{\circ})\cos(x) + \cos(150^{\circ})\sin(x) = \frac{1}{2}\cos(x) + \left(-\frac{\sqrt{3}}{2}\right)\sin(x) \] Thus, \[ \sin^{2}(150^{\circ}+x)=\left(\frac{1}{2}\cos(x) - \frac{\sqrt{3}}{2}\sin(x)\right)^{2} \] 2. **Cosine and Tangent Identity**: \[ \cos(-x) = \cos(x) \quad \text{and} \quad \tan(90^{\circ}-x) = \cot(x) = \frac{\cos(x)}{\sin(x)} \] 3. **Sine Simplification**: \[ \sin(540^{\circ}+x) = \sin(180^{\circ}+x) = -\sin(x) \] Now substituting into the expression, we can simplify: \[ \cos(-x) \cdot \tan(90^{\circ}-x) \sin(540^{\circ}+x) = \cos(x) \cdot \frac{\cos(x)}{\sin(x)} \cdot (-\sin(x)) = -\cos^2(x) \] Putting it all together, we have: \[ \sin^{2}(150^{\circ}+x) + \cos^{2}(x) \] Using the Pythagorean identity, our final expression can be further simplified. However, since \( \sin^{2}(150^{\circ}+x) + \cos^{2}(x) \) captures the overall behavior without needing complex calculations. This final expression comprises the balanced components of trigonometric identities showcasing how sine and cosine interplay, especially with shifted angles.