Given equations are:

$4a+kb+8=0$

$2a+2b+2=0$

Above equations are of the form

${{a}_{1}}a+{{b}_{1}}b-{{c}_{1}}=0$

${{a}_{2}}a+{{b}_{2}}b-{{c}_{2}}=0$

Now, ${{a}_{1}}=4,{{b}_{1}}=k,{{c}_{1}}=8$

${{a}_{2}}=2,{{b}_{2}}=2,{{c}_{2}}=2$

According to the question,

For unique solution, the condition is

${{a}_{1}}/{{a}_{2}}\ne {{b}_{1}}/{{b}_{2}}$

$4/2\ne k/2$

$\Rightarrow k\ne 4$

Therefore, the given equations will have unique solution for all real values of k other than $4$.