e) \( \left(-\frac{1}{2}+\frac{3}{4}\right)^{-1}-\sqrt[3]{-8} \cdot\left(-\frac{3}{2}\right)+(-2)^{2} \div(-2)^{3}-(-1)= \)

Let's break this down step by step! First, we simplify \( \left(-\frac{1}{2}+\frac{3}{4}\right)^{-1} \). This leads to \( \left(\frac{-2 + 3}{4}\right)^{-1} = \left(\frac{1}{4}\right)^{-1} = 4 \). Next, simplify \( \sqrt[3]{-8} \cdot \left(-\frac{3}{2}\right) \). The cube root of -8 is -2, so we have \( -2 \cdot \left(-\frac{3}{2}\right) = 3 \). For \( (-2)^{2} \div (-2)^{3} \), we get \( \frac{4}{-8} = -\frac{1}{2} \). Finally, the last part simplifies to \( -(-1) = 1 \). Putting it all together: \[ 4 - 3 - \frac{1}{2} + 1 \] Calculating that gives: \[ 4 - 3 + 1 - \frac{1}{2} = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \] So, the final result is \( \frac{3}{2} \).