$\left\{\begin{array}{l}y=x^{2}-2 x+1 \\ y=-x^{2}+3 x+4\end{array}\right.$

Step 1 ：First, we need to find the common points of the two functions, which means we need to solve the equation $x^{2}-2 x+1=-x^{2}+3 x+4$.

Step 2 ：Rearrange the equation, we get $2x^{2}-5x-3=0$.

Step 3 ：This is a quadratic equation, and we can solve it by using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, where $a=2$, $b=-5$, and $c=-3$.

Step 4 ：Substitute $a$, $b$, and $c$ into the formula, we get $x = \frac{5 \pm \sqrt{(-5)^{2}-4*2*(-3)}}{2*2}$.

Step 5 ：Simplify the equation, we get $x = \frac{5 \pm \sqrt{25+24}}{4}$.

Step 6 ：Further simplify the equation, we get $x = \frac{5 \pm \sqrt{49}}{4}$.

Step 7 ：So, the solutions are $x = \frac{5+7}{4}$ or $x = \frac{5-7}{4}$.

Step 8 ：Simplify the solutions, we get $x = 3$ or $x = -\frac{1}{2}$.

Step 9 ：Substitute $x = 3$ and $x = -\frac{1}{2}$ into the first equation $y=x^{2}-2 x+1$, we get $y = 3^{2}-2*3+1$ and $y = (-\frac{1}{2})^{2}-2*(-\frac{1}{2})+1$.

Step 10 ：Simplify the equations, we get $y = 4$ and $y = \frac{5}{4}$.

Step 11 ：So, the common points of the two functions are $(3,4)$ and $(-\frac{1}{2},\frac{5}{4})$.

Step 12 ：Check the solutions by substituting them into the second equation $y=-x^{2}+3 x+4$, we find that they satisfy the equation, so the solutions are correct.

Step 13 ：Therefore, the solutions to the system of equations are $\boxed{(3,4)}$ and $\boxed{(-\frac{1}{2},\frac{5}{4})}$.