### Solution:

#### Part (1): Proving the equation always has two distinct real roots

Given the quadratic equation $x^{2}+\left(m+3\right)x+m+1=0$, we need to examine the discriminant $\Delta$ to determine the nature of its roots. The discriminant $\Delta$ is given by:

$$\Delta = b^{2}-4ac$$

For our equation, $a=1$, $b=m+3$, and $c=m+1$. Substituting these values into the formula for $\Delta$, we get:

$$\begin{align*} \Delta &= (m+3)^{2}-4(m+1) \\ &= m^{2} + 6m + 9 - 4m - 4 \\ &= m^{2} + 2m + 5 \\ &= (m+1)^{2} + 4 \end{align*}$$

Since $(m+1)^{2} \geq 0$ for all real $m$, and adding $4$ to any non-negative number keeps it positive, we have $(m+1)^{2} + 4 > 0$ for all $m$. This means $\Delta > 0$, which implies the original equation always has two distinct real roots.

$\boxed{\text{The equation always has two distinct real roots.}}$

#### Part (2): Finding the value of $m$ and determining the roots

Given $|x_{1}-x_{2}|=2\sqrt{2}$, we can square both sides to eliminate the absolute value and simplify the equation:

$$\begin{align*} (x_{1}-x_{2})^{2} &= (2\sqrt{2})^{2} \\ (x_{1}+x_{2})^{2} - 4x_{1}x_{2} &= 8 \end{align*}$$

Using the relationships $x_{1}+x_{2}=-b/a=-\left(m+3\right)$ and $x_{1}x_{2}=c/a=m+1$, we substitute these into the equation:

$$\begin{align*} \left[-\left(m+3\right)\right]^{2} - 4(m+1) &= 8 \\ (m+3)^{2} - 4(m+1) &= 8 \\ m^{2} + 6m + 9 - 4m - 4 &= 8 \\ m^{2} + 2m - 3 &= 0 \end{align*}$$

Solving this quadratic equation for $m$, we find:

$$m_{1}=-3, \quad m_{2}=1$$

- When $m=-3$, the original equation becomes $x^{2}-2=0$. Solving this gives $x_{1}=\sqrt{2}, x_{2}=-\sqrt{2}$.
- When $m=1$, the original equation becomes $x^{2}+4x+2=0$. Solving this gives $x_{1}=-2+\sqrt{2}, x_{2}=-2-\sqrt{2}$.

$\boxed{m=-3 \text{ or } m=1}$

For $m=-3$, the roots are $\boxed{x_{1}=\sqrt{2}, x_{2}=-\sqrt{2}}$.

For $m=1$, the roots are $\boxed{x_{1}=-2+\sqrt{2}, x_{2}=-2-\sqrt{2}}$.