Let $A=\left(
\begin{array}{ccc}
1 & 1& 0 \\
0 & 1& 0 \\
0 &0 & 2 
\end{array}
\right),\ B=\left(
\begin{array}{ccc}
a & 1& 0 \\
b & 2& c \\
0 &0 & a+1 
\end{array}
\right)\ (a,\ b,\ c\in{\mathbb{C}}).$

(1) Find the condition for $a,\ b,\ c$ such that ${\text{rank} (AB-BA})\leq 1.$

(2) Under the condition of (1), find the condition for $a,\ b,\ c$ such that $B$ is diagonalizable.
1. **Finding the condition for \( a, b, c \) such that \(\text{rank}(AB - BA) \leq 1\):**

   First, we need to compute the commutator \([A, B] = AB - BA\).

   Given:
   \[
   A = \begin{pmatrix}
   1 & 1 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 2
   \end{pmatrix}, \quad
   B = \begin{pmatrix}
   a & 1 & 0 \\
   b & 2 & c \\
   0 & 0 & a+1
   \end{pmatrix}
   \]

   Compute \( AB \):
   \[
   AB = \begin{pmatrix}
   1 & 1 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 2
   \end{pmatrix}
   \begin{pmatrix}
   a & 1 & 0 \\
   b & 2 & c \\
   0 & 0 & a+1
   \end{pmatrix}
   = \begin{pmatrix}
   a + b & 3 & c \\
   b & 2 & c \\
   0 & 0 & 2(a+1)
   \end{pmatrix}
   \]

   Compute \( BA \):
   \[
   BA = \begin{pmatrix}
   a & 1 & 0 \\
   b & 2 & c \\
   0 & 0 & a+1
   \end{pmatrix}
   \begin{pmatrix}
   1 & 1 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 2
   \end{pmatrix}
   = \begin{pmatrix}
   a & a + 1 & 0 \\
   b & b + 2 & 2c \\
   0 & 0 & 2(a+1)
   \end{pmatrix}
   \]

   Now, compute \( AB - BA \):
   \[
   AB - BA = \begin{pmatrix}
   a + b & 3 & c \\
   b & 2 & c \\
   0 & 0 & 2(a+1)
   \end{pmatrix}
   - \begin{pmatrix}
   a & a + 1 & 0 \\
   b & b + 2 & 2c \\
   0 & 0 & 2(a+1)
   \end{pmatrix}
   = \begin{pmatrix}
   b & 2 - a & c \\
   0 & -b & -c \\
   0 & 0 & 0
   \end{pmatrix}
   \]

   For \(\text{rank}(AB - BA) \leq 1\), the matrix \(\begin{pmatrix} b & 2 - a & c \\ 0 & -b & -c \\ 0 & 0 & 0 \end{pmatrix}\) must have rank at most 1. This implies that the rows must be linearly dependent.

   From the second row, we see that \(b = 0\). Then the matrix simplifies to:
   \[
   \begin{pmatrix}
   0 & 2 - a & c \\
   0 & 0 & -c \\
   0 & 0 & 0
   \end{pmatrix}
   \]

   For the rank to be at most 1, we must have either \(2 - a = 0\) or \(c = 0\). Therefore, the conditions are:
   \[
   b = 0 \quad \text{and} \quad (a = 2 \text{ or } c = 0)
   \]

2. **Finding the condition for \( a, b, c \) such that \( B \) is diagonalizable under the condition \(\text{rank}(AB - BA) \leq 1\):**

   We consider the cases derived from the conditions \( b = 0 \) and \( a = 2 \text{ or } c = 0 \):

   - **Case 1: \( a = 2 \) and \( c \neq 0 \)**

     The characteristic polynomial of \( B \) is:
     \[
     \det(B - \lambda I) = \det \begin{pmatrix}
     2 - \lambda & 1 & 0 \\
     0 & 2 - \lambda & c \\
     0 & 0 & 3 - \lambda
     \end{pmatrix}
     = (2 - \lambda)^2 (3 - \lambda)
     \]

     The minimal polynomial is:
     \[
     m_B(\lambda) = (2 - \lambda)^2 (3 - \lambda)
     \]

     Since the minimal polynomial has a repeated root, \( B \) is not diagonalizable.

   - **Case 2: \( a = 2 \) and \( c = 0 \)**

     The characteristic polynomial of \( B \) is:
     \[
     \det(B - \lambda I) = \det \begin{pmatrix}
     2 - \lambda & 1 & 0 \\
     0 & 2 - \lambda & 0 \\
     0 & 0 & 3 - \lambda
     \end{pmatrix}
     = (2 - \lambda)^2 (3 - \lambda)
     \]

     The minimal polynomial is:
     \[
     m_B(\lambda) = (2 - \lambda)^2 (3 - \lambda)
     \]

     Since the minimal polynomial has a repeated root, \( B \) is not diagonalizable.

   - **Case 3: \( a \neq 2 \) and \( c = 0 \)**

     The characteristic polynomial of \( B \) is:
     \[
     \det(B - \lambda I) = \det \begin{pmatrix}
     a - \lambda & 1 & 0 \\
     0 & 2 - \lambda & 0 \\
     0 & 0 & a + 1 - \lambda
     \end{pmatrix}
     = (a - \lambda)(2 - \lambda)(a + 1 - \lambda)
     \]

     The minimal polynomial is:
     \[
     m_B(\lambda) = (a - \lambda)(2 - \lambda)(a + 1 - \lambda)
     \]

     Since the minimal polynomial has distinct roots, \( B \) is diagonalizable.

   Therefore, the conditions for \( B \) to be diagonalizable are:
   \[
   b = 0, \quad c = 0, \quad a \neq 2
   \]

The final answer is \( \boxed{ b = 0, \quad c = 0, \quad a \neq 2 } \)