Given 4. $m, n$ are positive integers, and in the polynomial $f(x)=(1+x)^{m}+(1+x)^{n}$, the coefficient of $x$ is 19.
1) Try to find the minimum value of the coefficient of $x^{2}$ in $f(x)$;
2) For the $m, n$ that make the coefficient of $x^{2}$ in $f(x)$ the smallest, find the term containing $x^{7}$ at this time.
Solve for the coefficient of $x$ in $f(x)=(1+x)^{m}+(1+x)^{n}$:
$$
\begin{array}{r}
C_{m}^{1}+C_{n}^{1}=m+n=19 . \\
\therefore \quad m=19-n .(1 \leqslant n \leqslant 18)
\end{array}
$$
1) The coefficient of $x^{2}$ in $f(x)$ is:
$$
\begin{aligned}
C_{m}^{2} & +C_{n}^{2}=\frac{1}{2}[m(m-1)+n(n \\
& -1)] \\
& =\frac{1}{2}[(19-n)(18-n) \\
& +n(n-1)] \\
& =\left(n-\frac{19}{2}\right)^{2}+\frac{323}{4} .
\end{aligned}
$$
$\because 1 \leqslant n \leqslant 18$, and $n \in N$,
$\therefore$ when $n=9$ or 10, the minimum value of $C_{m}^{2}+C_{n}^{2}$ is:
$$
C_{0}^{2}+C_{10}^{2}=81 \text {. }
$$
2) When the coefficient of $x^{2}$ in $f(x)$ is minimized,
$$
\left\{\begin{array} { l } 
{ m = 1 0 , } \\
{ n = 9 }
\end{array} \text { or } \left\{\begin{array}{l}
m=9, \\
n=10 .
\end{array}\right.\right.
$$

Thus, the coefficient of $x^{7}$ is $C_{0}^{7}+C_{10}^{7}=C_{0}^{6}+C_{10}^{3}=156$.