1. Given the hyperbola $C: \left(1-a^{2}\right) x^{2}+a^{2} y^{2}=a^{2}(a>1)$, let the vertex of the upper branch of the hyperbola be $A$, and the upper branch intersects the line $y=-x$ at point $P$. A parabola with focus at $A$, vertex at $M(0, m)$, and opening downwards passes through point $P$, and the slope of $P M$ is $k$ satisfying $\frac{1}{4} \leqslant k \leqslant \frac{1}{3}$. Find the range of the real number $a$.
1. The original equation can be transformed into $y^{2}-\frac{x^{2}}{a^{2} /\left(a^{2}-1\right)}=1$. Given $a>1$, we know $\frac{a^{2}}{a^{2}-1}>0$. Also, $A(0,1)$, so the equation of the parabola with focus at $A$ and vertex at $M(0, m)$, opening downwards, is $x^{2}=-4(m-1)(y-m)$. By solving $y=-x$ and $\left(1-a^{2}\right) x+a^{2} y^{2}=a^{2}$, we get $P(-a, a)$.

Since $P$ lies on the parabola, we have $a^{2}=-4(m-1)(a-m)$.(*) And $k_{M P}=\frac{m-a}{a}$, which gives $m=a k_{M P}+a$. Substituting this into (*) yields $4 a k_{M P}^{2}+4(a-1) k_{M P}-a=0$. Given $\frac{1}{4} \leqslant k_{M P} \leqslant \frac{1}{3}$ and $4 a>0$, the discriminant of the quadratic equation in $k_{M P}$, $\Delta=[4(a-1)]^{2}+4 \cdot 4 a \cdot a>0$, holds. Let $f(k)=4 a k^{2}+4(a-1) k-a$, and the axis of symmetry of this parabola is $k=$ $-\frac{4(a-1)}{2 \cdot 4 a}=\frac{1-a}{2 a}$. Since $a>1$, then $\frac{1-a}{2 a}0$ and $f\left(\frac{1}{4}\right) \cdot f\left(\frac{1}{3}\right) \leqslant 0$, i.e., $\left(4 a \cdot \frac{1}{16}+a-1-a\right) \cdot\left(4 a \cdot \frac{1}{9}+\frac{4 a-4}{3}-a\right) \leqslant 0$, i.e., $\left(\frac{1}{4} a-1\right)\left(\frac{7}{9} a-\frac{4}{3}\right) \leqslant 0$, hence $\frac{12}{7} \leqslant a \leqslant 4$ is the solution.