Step 1
We are given with following function

\[f\left( x \right) = 3{x^2} - x + 2\]

Step 2: f$(2)$
\[\begin{array}{l}f\left( 2 \right) = 3{\left( 2 \right)^2} - \left( 2 \right) + 2\\ = 3\left( 4 \right) - 2 + 2\\ = 12\end{array}\]Therefore,
\[f\left( 2 \right) = 12\]

Step 3: f$(-2)$
\[\begin{array}{l}f\left( { - 2} \right) = 3{\left( { - 2} \right)^2} - \left( { - 2} \right) + 2\\ = 3\left( 4 \right) + 2 + 2\\ = 16\end{array}\]Therefore,
\[f\left( { - 2} \right) = 16\]

Step 4: $f(a)$
\[\begin{array}{l}f\left( a \right) = 3{\left( a \right)^2} - \left( a \right) + 2\\ = 3{a^2} - a + 2\end{array}\]Therefore,
\[f\left( a \right) = 3{a^2} - a + 2\]

Step 5: $f(a)$
\[\begin{array}{l}f\left( { - a} \right) = 3{\left( { - a} \right)^2} - \left( { - a} \right) + 2\\ = 3{a^2} + a + 2\end{array}\]Therefore,
\[f\left( { - a} \right) = 3{a^2} + a + 2\]

Step 6: $f(a+1)$
\[\begin{array}{l}f\left( {a + 1} \right) = 3{\left( {a + 1} \right)^2} - \left( {a + 1} \right) + 2\\ = 3{a^2} + 3 + 6a - a - 1 + 2\\ = 3{a^2} + 5a + 4\end{array}\]Therefore,
\[f\left( {a + 1} \right) = 3{a^2} + 5a + 4\]

Step 7: $2f(a)$
\[\begin{array}{l}2f\left( a \right) = 2\left[ {3{a^2} - a + 2} \right]\\ = 6{a^2} - 2a + 4\end{array}\]Therefore,
\[2f\left( a \right) = 6{a^2} - 2a + 4\]

Step 8: $f(2a)$
\[\begin{array}{l}f\left( {2a} \right) = 3{\left( {2a} \right)^2} - 2a + 2\\ = 3\left( {4{a^2}} \right) - 2a + 2\\ = 12{a^2} - 2a + 2\end{array}\]Therefore,
\[f\left( {2a} \right) = 12{a^2} - 2a + 2\]

Step 9: $f(a^2)$
\[\begin{array}{l}f\left( {{a^2}} \right) = 3{\left( {{a^2}} \right)^2} - {a^2} + 2\\ = 3\left( {{a^4}} \right) - {a^2} + 2\\ = 3{a^4} - {a^2} + 2\end{array}\]Therefore,
\[f\left( {{a^2}} \right) = 3{a^4} - {a^2} + 2\]

Step 10: $f(a^2)$
\[\begin{array}{l}{\left[ {f\left( a \right)} \right]^2} = {\left( {3{a^4} - {a^2} + 2} \right)^2}\\ = \left( {9{a^4} + {a^2} + 4 - 2 \times 3{a^2} \times a - 2 \times a \times 2 + 2 \times 2 \times 3{a^2}} \right)\\ = 9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\end{array}\]Therefore,
\[{\left[ {f\left( a \right)} \right]^2} = 9{a^4} - 6{a^3} + 13{a^2} - 4a + 4\]

Step 11: $f(a+h)$
\[\begin{array}{l}f\left( {a + h} \right) = 3{\left( {a + h} \right)^2} - \left( {a + h} \right) + 2\\ = 3\left( {{a^2} + {h^2} + 2ah} \right) - a - h + 2\\ = 3{a^2} + 6ah + 3{h^2} - a - h + 2\end{array}\]Therefore,
\[f\left( {a + h} \right) = 3{a^2} + 6ah + 3{h^2} - a - h + 2\]