We have,
$$
f(x)=\left\{\begin{array}{cc}
a x+b, & x \leq-1 \\
2 x^2+2 b x-\frac{a}{2}, & -1 < x < 1 \\
7, & x \geq 1
\end{array}\right.
$$
$f(x)$ is continuous on $R$.
$\therefore f(x)$ is continuous at $x=-1$ and $x=1$ at $x=-1$
$$
\begin{aligned}
& \lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x) \\
& \Rightarrow \quad a(-1)+b=2(-1)^2+2 b(-1)-\frac{a}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad-a+b=2-2 b-\frac{a}{2} \\
& \Rightarrow \quad-\frac{a}{2}+3 b=2 \\
& \text { at } \quad x=1 \\
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\
& \Rightarrow 2(1)^2+2 b(1)-\frac{a}{2}=7 \\
& \Rightarrow \quad-\frac{a}{2}+2 b=5
\end{aligned}
$$

From Eqs. (i) and (ii), we get
$$
b=-3
$$

Substitute value of $b$ in Eq. (i)
$$
\begin{aligned}
-\frac{a}{2}+3(-3) & =2 \quad-\frac{a}{2}=11 \\
a & =-22
\end{aligned}
$$

Hence, $\quad(a, b)=(-22,-3)$