Let $M$ be a non-zero $3\times 3$ matrix satisfying $M^3=O$

$M^3=0$ implies $\det(M-xI)=-x^3$ (since $M$ satisfies its own characteristic equation). Decompose expression in question into linear factors: $$\det(\frac 1 2M^2\pm M+I)=\frac 1 8\det(M-x_{\pm1}I)\det(M-x_{\pm2}I)=\frac 1 8x_{\pm1}^3x_{\pm2}^3=\frac 1 82^3=1$$ where $x_{\pm}1,x_{\pm2}$ are roots of $x^2\pm2x+2=0$ and hence $x_{\pm1}x_{\pm2}=2$.

All of above is really not necessary $$\left(\frac{M^2}{2}+M+I\right)\left(\frac{M^2}{2}-M+I\right)=I$$

It is easy from here onwards

If we put $M$ in Jordan canonical form, for some invertible matrix $Q$ we will have $$M=QAQ^{-1}$$ where $$A=\left(\begin{smallmatrix}0&a&0\\0&0&b\\0&0&0\end{smallmatrix}\right)$$ where $a,b$ are each $0$ or $1$ (not both $0$ since $M\neq 0$). Hence, we write $$0.5M^2+M+I=0.5QAQ^{-1}QAQ^{-1}+QAQ^{-1}+QIQ^{-1}=$$ $$=Q(0.5A^2+A+I)Q^{-1}$$ Taking determinants, we get $$det(Q)det(0.5A^2+A+I)det(Q)^{-1}=det(0.5A^2+A+I)=1$$ since $$0.5A^2+A+I=\left(\begin{smallmatrix}1&a&\frac{ab}{2}\\0&1&b\\0&0&1\end{smallmatrix}\right)$$

The other three questions are similar.