### Solution:

#### Part (Ⅰ)

To solve the quadratic equation $x^{2}+6x+5=0$, we factor it:

$$x^{2}+6x+5 = (x+5)(x+1) = 0$$

This gives us two possible equations:

1. $x + 5 = 0 \Rightarrow x = -5$
2. $x + 1 = 0 \Rightarrow x = -1$

Thus, the solutions are $x_{1} = -5$ and $x_{2} = -1$. Therefore, we have:

$$\boxed{x_{1} = -5, x_{2} = -1}$$

#### Part (Ⅱ)

Given the quadratic equation $x^{2}-3x+k=0$ has two real roots $x_{1}$ and $x_{2}$, and $(x_{1}-1)(x_{2}-1)=-6$, we use the properties of quadratic equations:

1. Sum of roots: $x_{1} + x_{2} = 3$
2. Product of roots: $x_{1}x_{2} = k$

From the given condition $(x_{1}-1)(x_{2}-1)=-6$, we expand and substitute the known sum and product of roots:

$$x_{1}x_{2} - (x_{1} + x_{2}) + 1 = -6$$

Substituting $x_{1} + x_{2} = 3$ and $x_{1}x_{2} = k$:

$$k - 3 + 1 = -6 \Rightarrow k = -4$$

With $k = -4$, the equation becomes $x^{2} - 3x - 4 = 0$, which factors to:

$$(x-4)(x+1) = 0$$

This gives us two possible equations:

1. $x - 4 = 0 \Rightarrow x = 4$
2. $x + 1 = 0 \Rightarrow x = -1$

Thus, the solutions are $x_{1} = 4$ and $x_{2} = -1$. Therefore, we have:

$$\boxed{k = -4, x_{1} = 4, x_{2} = -1}$$