1. Given the function $f(x)=x^{3}+a x^{2}+b x+a^{2}$ has an extremum of 10 at $x=1$. Then the real number pair $(a, b)$ is $\qquad$ .
II. 1. $(4,-11)$.
Given that the function has an extremum of 10 at $x=1$, we have
$$
\left\{\begin{array}{l}
f^{\prime}(1)=0, \\
f(1)=10,
\end{array} \text { i.e., } \left\{\begin{array}{l}
3+2a+b=0, \\
1+a+b+a^{2}=10 .
\end{array}\right.\right.
$$

Solving, we get $\left\{\begin{array}{l}a_{1}=4, \\ b_{1}=-11 ;\end{array}\left\{\begin{array}{l}a_{2}=-3, \\ b_{2}=3 .\end{array}\right.\right.$
Substituting $(4,-11)$, we have
$$
\begin{array}{l}
f(x)=x^{3}+4 x^{2}-11 x+16, \\
f^{\prime}(x)=3 x^{2}+8 x-11=(x-1)(3 x+11) .
\end{array}
$$

It can be concluded that $x=1$ is a point of local minimum with a value of 10.
Substituting $(-3,3)$, we have
$$
\begin{array}{l}
f(x)=x^{3}-3 x^{2}+3 x+9, \\
f^{\prime}(x)=3 x^{2}-6 x+3=3(x-1)^{2} \geqslant 0 .
\end{array}
$$

Thus, for $x \in \mathbf{R}$, $f(x)$ does not decrease, and $x=1$ is not an extremum point.
In summary, $(a, b)$ is $(4,-11)$.
Note: The $x$ values where $f^{\prime}(x)=0$ are only suspected points of extremum, and the sufficiency needs to be verified.