The absolute maximum value of $f(x)$ can be achieved as $\frac{13}{2}$ when $x=0$. Therefore, maximum value $2b$ can only be one of $f(a)$, $f(b)$, or $f(0)$. Correspondingly, minimal value $2a$ can be either $f(a)$ or $f(b)$. $\underline{case: a \le 0 < b}$ Then maximum value $2b=\frac{13}{2}\implies b=\frac{13}{4}$. It follows that minimal value $2a = f(a)$ or $f(b)$. - If $2a=f(b)$, then $2a=f(\frac{13}{4})=\frac{39}{32}\implies a =\frac{39}{64} > 0$. This contradicts to the assumption $a \le 0 < b$. - If $2a = f(a)$, then $2a = -\frac{a^2}{2}+\frac{13}{4}\implies a =-2-\sqrt{17}$. Hence $(a, b)=\Big(-2-\sqrt{17}, \dfrac{13}{4}\Big)$ is one possible solution. $\underline{case: a \le b \le 0}$ Then $f(x)$ monotonically increase in range $[a, b]$. Therefore we must have $$ \left\{ \begin{array}{ccc} 2a &= f(a)\\ \\ 2b &= f(b) \end{array} \right. \implies \left\{ \begin{array}{ccc} 2a &= -\frac{a^2}{2} + \frac{13}{2}\\ \\ 2b &= -\frac{b^2}{2} + \frac{13}{2} \end{array} \right. $$ This means $a$ and $b$ are two roots of equation $$\frac{x^2}{2}+2x-\frac{13}{2}=0$$ By Vieta's theorem, $a\cdot b = -13$ which means they have different signs. This contradicts to the assumption $a< b< 0$. $\underline{case: 0\le a < b}$ Then $f(x)$ monotonically decreases in range $[a, b]$. Hence $$ \left\{ \begin{array}{ccc} 2a &= f(b)\\ \\ 2b &= f(a) \end{array} \right. \implies \left\{ \begin{array}{ccc} 2a &= -\frac{b^2}{2} + \frac{13}{2}\\ \\ 2b &= -\frac{a^2}{2} + \frac{13}{2} \end{array} \right. \implies \left\{ \begin{array}{ccc} a &= 1\\ \\ b &= 3 \end{array} \right. $$ Therefore, in conclusion, the answer is $(a, b) = \Big(-2-\sqrt{17}, \dfrac{13}{4}\Big)$ or $(1, 3)$.