I didn't have any idea about the question. First, I asked ChatboxGPT and got the hint, then wrote the below solution for (5-1).
\begin{array}{l} \because q=Ap+B\Longrightarrow B=q-Ap\\ \therefore A^{2} +B^{2} =A^{2} +( q-Ap)^{2} =A^{2} +q^{2} -2Apq+A^{2} p^{2}\\ \frac{d}{dA}\left( A^{2} +B^{2}\right) =2A-2pq+2p^{2} A,\ \frac{d^{2}}{dA^{2}}\left( A^{2} +B^{2}\right) =2+2p^{2} >0\\ \therefore let\ \frac{d}{dA}\left( A^{2} +B^{2}\right) =0\ would\ get\ the\ minimum\ of\ A\\ 2A-2pq+2p^{2} A=0\\ 2A\left( 1+p^{2}\right) =2pq\\ A_{min} =\frac{pq}{1+p^{2}}\\ \therefore B_{min} =q-A_{min} p=q-\frac{pq}{1+p^{2}} =\frac{q}{1+p^{2}}\\ \therefore A_{min}^{2} +B_{min}^{2} =\left(\frac{pq}{1+p^{2}}\right)^{2} +\left(\frac{q}{1+p^{2}}\right)^{2} =\frac{p^{2} q^{2} +q^{2}}{\left( 1+p^{2}\right)^{2}} \end{array}
I'm confused about getting the minimum value of $A^{2} +B^{2}$ through the minimum value of A. When A is minimum, B(y-intercept) is maximum. Is it correct to obtain the minimum value? If it is correct, what about through the minimum value of B to get the answer? (I'm sorry I don't have reference answer)
For (5-2), here is my idea. \begin{array}{l} \because |a+b|\leqslant |a|+|b|\\ \therefore ( |A|+|B|)_{min} =\sqrt{( A+B)^{2}} =\sqrt{A^{2} +B^{2} +2AB} \end{array} Please tell me if it is correct. Thanks for your help.
Solution (5-2):
Let $G(A,B)\mapsto |A|+|B|$.
Then $G_{min}(A,p,q)= G(A,q-A p) = |A|+|q-A p|$.
HINT: The $|x|$ function has one global minimum at $x=0$.
Hence $G_{min}(A,p,q)$ may be minimal at $G_{min}(0,p,q)=|q|$ or $G_{min}(\frac{q}{p},p,q)=|\frac{q}{p}|$, whoever is smaller.
Finally we get the minimum value at $min(|\frac{q}{p}|,|q|)$.