(1)

\begin{align*} a^{\left(1\right)}\left(b^{\left(1\right)}c\right) & =a+b+c\\ & =\left(a^{\left(1\right)}b\right)^{\left(1\right)}c \end{align*} \begin{align*} a^{\left(2\right)}\left(b^{\left(2\right)}c\right) & =abc\\ & =\left(a^{\left(2\right)}b\right)^{\left(2\right)}c \end{align*}

(2)

\(n=0\)のとき、

\begin{align*} a^{\left(0\right)}\left(b^{\left(0\right)}c\right) & =a^{\left(0\right)}\left(c+1\right)\\ & =c+2 \end{align*} \begin{align*} \left(a^{\left(0\right)}b\right)^{\left(0\right)}c & =c+1 \end{align*} となるので、
\[ a^{\left(0\right)}\left(b^{\left(0\right)}c\right)\ne\left(a^{\left(0\right)}b\right)^{\left(0\right)}c \]

\(n=2,3,\cdots\)のとき、

\(c=0\)とする。
\begin{align*} a^{\left(n\right)}\left(b^{\left(n\right)}0\right) & =a^{\left(n\right)}1\\ & =a \end{align*} \[ \left(a^{\left(n\right)}b\right)^{\left(n\right)}0=1 \] となり、
\[ a^{\left(n\right)}\left(b^{\left(n\right)}0\right)\ne\left(a^{\left(n\right)}b\right)^{\left(n\right)}0 \] となるので、一般に
\[ a^{\left(n\right)}\left(b^{\left(n\right)}c\right)\ne\left(a^{\left(n\right)}b\right)^{\left(n\right)}c \] となる。