a: f(0)=3

=>\(a\cdot0+b=3\)

=>b=3

=>f(x)=ax+3

f(2)=7

=>2a+3=7

=>2a=4

=>a=2

=>f(x)=2x+3

b: f(2)=8

=>\(a\cdot2+b=8\)

=>b=-2a+8

f(-2)=12

=>\(a\cdot\left(-2\right)+b=12\)

=>-2a+b=12

=>-2a-2a+8=12

=>-4a=4

=>a=-1

=>\(b=-2a+8=-2\cdot\left(-1\right)+8=2+8=10\)

Vậy: f(x)=-x+10

Ta có : \(f\left(1\right)=g\left(2\right)\)

\(\Rightarrow2^2+a+8=1^2-5-b\)

\(\Rightarrow a+8=-4-b\)

\(\Rightarrow a+b=-12\)(1)

Mặt khác : \(f\left(-1\right)=g\left(5\right)\)

\(\Rightarrow\left(-2\right)^2-a+4=5^2-5.5-b\)

\(\Rightarrow8-a=-b\)

\(\Rightarrow a=8+b\)(2)

Thay (2) vào (1), ta có : \(8+2b=12\)

\(\Rightarrow2b=4\)

\(\Rightarrow b=2\)(3)

Thay (3) vào (2), ta có : \(a=8+2=10\)

Vậy a = 10 ; b = 2

Ta có: \(\orbr{\begin{cases}f\left(1\right)=a+b\\f\left(-2\right)=-2a+b\end{cases}}\)

Tương tự: \(\orbr{\begin{cases}g\left(2\right)=3\\g\left(1\right)=1\end{cases}}\)

Mặt khác \(f\left(1\right)=g\left(2\right)\)và \(f\left(-2\right)=g\left(1\right)\)

Suy ra: \(\hept{\begin{cases}a+b=3\\-2a+b=1\end{cases}\Rightarrow a+b=-2a+b+2\Rightarrow a=-2a+2\Leftrightarrow a=\frac{2}{3}}\)

Suy ra: \(b=\frac{7}{3}\)

\(F\left(1\right)=a+b=3;F\left(-2\right)=-2a+b=2\)

\(\Rightarrow a=\dfrac{1}{3};b=\dfrac{8}{3}\)

cảm giác hơi tắt lm sao ấy