$a)\{\begin{array}{l}{S}_3=35\\ a_1-a_4=\frac{35}{2}\end{array}$  

$a_1-a_1{q}^3=\frac{35}{2}$  

$a_1\left(1-q^3\right)=\frac{35}{2}$  

$a_1=\frac{35}{2\left(1-q^3\right)}$  

 

$S_3=a_1\cdot \frac{1-q^3}{1-q}$  

$\frac{35}{2\left(1-q^3\right)}\cdot \frac{1-q^3}{1-q}=35\mid :35$  

$\frac{1}{2-2q}=1$  

$1=2-2q$  

$-2q=-1$  

$q=\frac{1}{2}$  

$a_1=\frac{35}{2\left(1-\frac{1}{8}\right)}=\frac{35}{2\cdot \frac{7}{8}}=\frac{35}{\frac{7}{4}}=35\cdot \frac{4}{7}=5\cdot 4=20$  

 

$a_n=a_1\cdot q^{n-1}=20\cdot {\left(\frac{1}{2}\right)}^{n-1}=5\cdot 2^2\cdot {\left(2^{-1}\right)}^{n-1}=5\cdot 2^{2-\left(n-1\right)}=5\cdot 2^{3-n}$  

 

$b)\{\begin{array}{l}{S}_4=20\\ a_1-a_3=8\end{array}$  

 

$a_1-a_3=8$  

$a_1-a_1{q}^2=8$  

$a_1\left(1-q^2\right)=8$  

$a_1=\frac{8}{1-q^2}$  

 

$S_4=a_1\cdot \frac{1-q^4}{1-q}$  

$\frac{8}{1-q^2}\cdot \frac{1-q^4}{1-q}=20$  

$\frac{8}{1-q^2}\cdot \frac{\left(1-q^2\right)\left(1+q^2\right)}{1-q}$  

$\frac{8\left(1+q^2\right)}{1-q}=20$  

$8+8q^2=20-20q$  

$8q^2+20q-12=0$  

$2q^2+5q-3=0$  

$2q^2+6q-q-3=0$  

$2q\left(q+3\right)-\left(q+3\right)=0$  

$\left(q+3\right)\left(2q-1\right)=0$  

$q_1=-3\vee q_2=\frac{1}{2}$  

 

$\{\begin{array}{l}{q}_1=-3\\ a_1=\frac{8}{1-9}\end{array}\vee \{\begin{array}{l}{q}_2=\frac{1}{2}\\ a_1=\frac{8}{1-\frac{1}{4}}\end{array}$  

$\{\begin{array}{l}{q}_1=-3\\ a_1=-1\end{array}\vee \{\begin{array}{l}{q}_2=\frac{1}{2}\\ a_1=\frac{32}{3}\end{array}$  

$a_n=-{\left(-3\right)}^{n-1}\vee a_n=\frac{32}{3}\cdot {\left(\frac{1}{2}\right)}^{n-1}$  

$a_n=-{\left(-3\right)}^{n-1}\vee a_n=\frac{1}{3}\cdot 2^5\cdot 2^{1-n}$  

$a_n=-{\left(-3\right)}^{n-1}\vee a_n=\frac{1}{3}\cdot 2^{6-n}$