1) $$f(3) = \frac{1}{3} \cdot 3^2 + 2 \cdot 3 = \frac{1}{3} \cdot 9 + 6 = 3 + 6 = 9$$
$$f(-1) = \frac{1}{3} \cdot (-1)^2 + 2 \cdot (-1) = \frac{1}{3} \cdot 1 - 2 = \frac{1}{3} - \frac{6}{3} = -\frac{5}{3}$$
Ответ: $$f(3)=9$$, $$f(-1)=-\frac{5}{3}$$