Знатоки алгебры сюда!

1)
$ax^2+3ax+6=0;\\ x_1^2+x_2^2=1;\\ D=9a^2-24a;\\ x_1=\frac{-3a-\sqrt{9a^2-24a}}{2a};\\ x_2=\frac{-3a+\sqrt{9a^2-24a}}{2a};\\ x_1^2+x_2^2=1=\left(\frac{-3a-\sqrt{9a^2-24a}}{2a}\right)^2+\left(\frac{-3a+\sqrt{9a^2-24a}}{2a}\right)^2=\\ =\frac{9a^2+6a\sqrt{9a^2-24a}+9a^2-24a+9a^2-6a\sqrt{9a^2-24}+9a^2-24a}{4a^2}=\\ =\frac{36a^2-48a}{4a^2}=9-\frac{12}{a}=1\\ 9-1=\frac{12}{a};\\ 8=\frac{12}{a};\\ \frac{a}{12}=\frac18;\\ a=\frac{12}{8}=\frac32\\ a=1,5;\\$
проверим
$\frac32x^2+3\cdot\frac32x+6=0;\\ \frac32x^2+\frac92x+6=0;\\ D=\frac{81}{4}-4\cdot\frac32\cdot6=\frac{81}{4}-36=\frac{81-144}{4}=-\frac{63}{4};\\ x_1=\frac{-\frac92-i\sqrt{\frac{63}{4}}}{2\cdot\frac32}=\frac{-9-i\sqrt{63}}{6};\\ x_2=\frac{-\frac92+i\sqrt{\frac{63}{4}}}{2\cdot\frac32}=\frac{-9+i\sqrt{63}}{6};\\ 1=x_1^2+x_2^2=\frac{81+18i\sqrt{63}-63+81-18i\sqrt{63}-63}{36}=\frac{2\cdot81-2\cdot63}{36}=\\ =\frac{2(81-63)}{36}=\frac{2\cdot18}{36}=\frac{36}{36}=1$
Значит $a=\frac32$

2)
$4x^2-7x+1=0;\\ (1+x_1)(1+x_2)-?\\ teorema Vietta:\\ 1+x_1+x_2+x_1\cdot x_2=1+(x_1+x+2)+x_1\cdot x_2\\ 4(x^2-\frac74+\frac14)=0;\\ x_1+x_2=\frac74;\\ x_1\cdot x_2=\frac14;\\ 1+(x_1+x_2)+x_1\cdot x_2=1+\frac74+\frac14=1+\frac84=1+2=3;\\ (1+x_1)\cdot(1+x_2)=3.$