(a) Yes, your mistake is with the elastic force, which you have written as $k\Delta L$. This is the tension $T$ in each of the springs, but it is not the force pulling each mass towards the centre.

There are 2 tension forces $T$ acting on each mass $m$. The component of each in the direction of the centre is $T\cos45^{\circ}=\frac{1}{\sqrt2}T$. Therefore the total elastic force on each mass towards the centre is $\frac{2}{\sqrt2}T=\sqrt2 T$.

Writing $F=ma$ with centripetal acceleration $a=\omega^2x$ we have $$\sqrt2 T=m\omega^2 x$$ $$\sqrt2 k\Delta L=m\omega^2(\frac{L+\Delta L}{\sqrt2})$$ $$2k\Delta L=m\omega^2(L+\Delta L)$$ $$\Delta L=\frac{m\omega^2L}{2k-m\omega^2}$$

(b) The above equation tells us that there is no equilibrium position for $2k \lt m\omega^2$, because $\Delta L$ would then be -ve, so the spring force would be outward - ie there would be no centripetal (inward) force. But it does not tell us if any equilibrium position is stable.

As you found out, to do that we have to examine how the resultant force on the mass changes as we increase/decrease the radius $x$ close to the equilibrium position $x_0$. If the resultant force becomes -ve (inward) for $x \gt x_0$ and +ve (outward) for $x \lt x_0$ then the equilibrium is stable.

Mathematically, we need to examine $\frac{dF}{dx}$ at the equilibrium position $x_0$. If it is $\lt 0$ - ie resultant force decreasing as we move outwards - then the equilibrium position is stable.

In the rotating frame of reference the resultant force $F(x)$ on each mass $m$ is the sum of the outward (+ve) centrifugal force and the elastic central force, which can be inward (-ve) or outward (+ve). Therefore : $$F(x)=m\omega^2 x-2k(x-\frac{L}{\sqrt2})$$ in which I have written the extension $\Delta L$ in terms of $x=\frac{L+\Delta L}{\sqrt2}$, and $F(x)$ is measured in the outward direction, the same as $x$.

Now the question has been somewhat ambiguous about $\omega$, so there are 2 cases to consider. We are not told whether (i) $\omega$ is fixed - eg regulated by an external torque (a **governor**) which keeps $\omega$ constant regardless of whether the masses move inwards or outwards, or (ii) after $\omega$ is given a fixed value at an equilibrium position the regulator is disconnected - then it is the **angular momentum** $m\omega x^2$ which is constant.

**(i) angular velocity $\omega$ is constant**

$$\frac{dF}{dx}=m\omega^2-2k$$ We already know that we must have $2k \gt m\omega^2$ for all equilibrium positions, so in this case $\frac{dF}{dx} \lt 0$ for all equilibrium positions, hence **all equilibrium positions are stable**.

**(ii) angular momentum $m\omega x^2$ is constant**

The angular velocity and radius at the equilibrium position are $\omega_0, x_0$. If the mass moves to a different radius $x$ the angular velocity $\omega$ changes but the angular momentum remains constant : $m\omega x^2 = m \omega_0 x_0^2$. Then $m\omega^2 x=\frac{m\omega_0^2 x_0^4}{x^3}$. Therefore : $$F(x)=\frac{m\omega_0^2 x_0^4}{x^3}-2k(x-\frac{L}{\sqrt2})$$ $$\frac{dF}{dx}=-\frac{3m\omega_0^2 x_0^4}{x^4}-2k$$ which is clearly $\lt 0$ for all values of $x$. So in this case also we find that **all equilibrium positions are stable**.

The only condition for stable equilibrium is $2k \gt m\omega^2$.