Well, of course, any qubit gate that you apply must be an 8x8 unitary matrix. For example, if you want to apply an H gate to the second bit, you are applying the unitary transformation
$$\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 0 & 1 & 0 &0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 &0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 &0 & 0 & 0 & 0 \\
0 & 1 & 0 & -1 &0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 &1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 &0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 &1& 0 & -1 & 0 \\
0 & 0 & 0 & 0 &0 & 1 & 0 & -1 \\
\end{pmatrix}
$$
But work with 8x8 matrices is very tedious, so it's better to always work in tensor product, in this case multiply by the previous matrix is equivalent to do
$$H_2\left|000\right> = \left(I\left|0\right>\right)\left(H\left|0\right>\right)\left(I\left|0\right>\right) = \left(\left|0\right>\right)\left(\frac{\left|0\right>+\left|1\right>}{\sqrt{2}}\right)\left(\left|0\right>\right) = \frac{\left|000\right>+\left|010\right>}{\sqrt{2}}$$
Or, if you want to apply an ##X## gate to the third qubit you simply do
$$X_3\left|000\right> = \left(I\left|0\right>\right)\left(I\left|0\right>\right)\left(X\left|0\right>\right) = \left(\left|0\right>\right)\left(\left|0\right>\right)\left(\left|1\right>\right)
=\left|001\right>$$
which is equivalent to multiply by the matrix
$$\begin{pmatrix}
0 & 1 & 0 & 0 &0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 &0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 &0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 &0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 &0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 &1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 &0& 0 & 0 & 1 \\
0 & 0 & 0 & 0 &0 & 0 & 1 & 0 \\
\end{pmatrix}
$$
But we all will agree that is better to work qubit by qubit. The CNOT gate works in the same way. Now, try to apply some qubit gates to go from ##\left|000\right>## to ##\frac{\left|000\right>+\left|111\right>}{\sqrt{2}}##