$\text{a)}\\f(x)=ax+b\\f(-1)=2 \to -a+b=2\to b=a+2\,(1)\\f(3)=-1\to 3a+b=-1\to b=-1-3a\,(2)\\\text{Từ }(1),\,(2) \to a+2=-1-3a\,(=b)\to 4a=-3 \to a=-\dfrac{3}{4}\to b=-\dfrac{3}{4}+2=\dfrac{5}{4}\\\text{Vậy }a=\dfrac{3}{4};\,b=\dfrac{5}{4}\\\text{b)}\\g(x)=5x^2+bx+c\\g(2)=5\to 5.2^2+2b+c=5\to 2b+c=-15\to c=-2b-15\,(1)\\g(1)=-1\to 5+b+c=-1\to c=-b-6\,(2)\\\text{Từ }(1),\,(2)\to -2b-15=-b-6\,(=c)\\\to b=-9\to c=-(-9)-6=3\\\text{Vậy }b=-9;\,c=3$