Lower Heating value calculation from chemcial combustion equation of Butane
I see three flaws in your reasoning. 

First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.

Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables. 

$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$

Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:

$$\ce{C4H8 + 6O2 -&gt; 4CO2 + 4H2O}$$

To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.

$$\ce{C4H8 + (6-6n_1)O2 -&gt; n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$ 

Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.