### Solution:

#### Part (1): Proof

Given $\sin C\sin(A-B) = \sin B\sin(C-A)$, we start by expanding the sine of differences:
$$\begin{align*} \sin C(\sin A\cos B - \cos A\sin B) &= \sin B(\sin C\cos A - \cos C\sin A) \\ \sin A\sin B\cos C + \sin A\cos B\sin C &= \sin B\sin C\cos A + \sin A\sin B\sin C \\ \sin A(\sin B\cos C + \cos B\sin C) &= 2\cos A\sin B\sin C \\ \sin A\sin(B + C) &= 2\cos A\sin B\sin C \end{align*}$$
Since $B + C = 180^\circ - A$ and $\sin(180^\circ - A) = \sin A$, we have:
$$\begin{align*} \sin^2 A &= 2\cos A\sin B\sin C \end{align*}$$
Applying the Law of Sines, $a/sinA = b/sinB = c/sinC$, and the Law of Cosines, $a^2 = b^2 + c^2 - 2bc\cos A$, we find:
$$\begin{align*} a^2 &= 2bc\cos A \\ 2a^2 &= b^2 + c^2 \end{align*}$$
Thus, we have proved that $2a^2 = b^2 + c^2$.

#### Part (2): Finding the Perimeter

Given $a = 5$ and $\cos A = \frac{25}{31}$, we calculate:
$$\begin{align*} b^2 + c^2 &= 2a^2 = 2 \times 5^2 = 50 \\ 2bc &= \frac{a^2}{\cos A} = \frac{25}{\frac{25}{31}} = 31 \end{align*}$$
Thus, the sum of squares and twice the product gives:
$$\begin{align*} (b + c)^2 &= b^2 + c^2 + 2bc = 50 + 31 = 81 \\ b + c &= \sqrt{81} = 9 \end{align*}$$
Therefore, the perimeter of $\triangle ABC$ is:
$$\begin{align*} a + b + c &= 5 + 9 = 14 \end{align*}$$
Hence, the perimeter is $\boxed{14}$.

### Solution:

#### Part (1): Proving $2a^{2}=b^{2}+c^{2}$

Given that $\sin C\sin (A-B)=\sin B\sin (C-A)$, we can expand both sides using trigonometric identities:
$$\begin{align*} \sin C(\sin A\cos B-\cos A\sin B) &= \sin B(\sin C\cos A-\cos C\sin A) \\ \Rightarrow \sin A\sin B\cos C + \sin A\cos B\sin C &= \sin B\sin C\cos A + \sin B\cos C\sin A \\ \Rightarrow \sin A(\sin B\cos C + \cos B\sin C) &= 2\cos A\sin B\sin C \\ \Rightarrow \sin A\sin (B+C) &= 2\cos A\sin B\sin C. \end{align*}$$

Using the Law of Sines, $a/sinA = b/sinB = c/sinC$, we can express $a^2$ in terms of $b$, $c$, and $\cos A$:
$$\begin{align*} a^{2} &= 2bc\cos A. \end{align*}$$

Applying the Law of Cosines, $a^{2}=b^{2}+c^{2}-2bc\cos A$, and substituting $2bc\cos A$ from the previous step, we get:
$$\begin{align*} a^{2} &= b^{2}+c^{2}-2bc\cos A \\ 2a^{2} &= b^{2}+c^{2}. \end{align*}$$

Therefore, we have proved that $2a^{2}=b^{2}+c^{2}$.

#### Part (2): Finding the Perimeter when $a=5$ and $\cos A=\frac{25}{31}$

Given $a=5$ and $\cos A=\frac{25}{31}$, we calculate $b^{2}+c^{2}$ and $2bc$ as follows:
$$\begin{align*} b^{2}+c^{2} &= 2a^{2} = 2\times 5^{2} = 50, \\ 2bc &= \frac{a^{2}}{\cos A} = \frac{25}{\frac{25}{31}} = 31. \end{align*}$$

To find $b+c$, we calculate $(b+c)^{2}$:
$$\begin{align*} (b+c)^{2} &= b^{2}+c^{2}+2bc = 50 + 31 = 81, \\ \Rightarrow b+c &= \sqrt{81} = 9. \end{align*}$$

Therefore, the perimeter of $\triangle ABC$ is:
$$\begin{align*} a+b+c &= 5 + 9 = 14, \\ &= \boxed{14}. \end{align*}$$