Three, let the roots of the equation $x^{2}+p x+q$ be 1 greater than the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$, and the difference between the roots of the equation $x^{2}+p x+q=0$ is equal to the difference between the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$, find the solutions of these two equations.
Three, Solution: Let the roots of the equation $x^{2}+2 q x+\frac{1}{2} p=0$ be $\alpha, \beta$. Then the roots of the equation $x^{2}+p x+q=0$ are $\alpha+1, \beta+1$.
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
\alpha+\beta=-2 q, \\
\alpha \beta=\frac{1}{2} p, \\
(\alpha+1)+(\beta+1)=-p, \\
(\alpha+1)(\beta+1)=q .
\end{array}\right.
$$

From (2) and (3), we get
$$
2 \alpha \beta+\alpha+\beta+2=0,
$$

From (1) and (4), we get
$$
2 \alpha \beta+3\left(\alpha+\beta\right)+2=0,
$$

From (3) and (6), we get
$$
\alpha+\beta=0, \alpha \beta=-1 .
$$

Therefore, the values of $\alpha, \beta$ are $\pm 1$.
$\therefore$ The solutions of the equation $x^{2}+p x+q=0$ are 0, 2; the solutions of the equation $x^{2}+2 q x+\frac{1}{2} p=0$ are $\pm 1$.