Convolution of Input signal and Impulse response
In the second case, which caused you the trouble, you should first use the sifting property of the impulse to find out the input signal $x[n]$ and then proceed with convolution.

sifting property is: 
$$ f[n] \delta[n-d] = f[d]\delta[n-d]$$

Apply this to simplify your input signal $x[n]$ as follows:

\begin{align}
x[n] &amp;= (\frac{1}{4})^{-n} \left(\delta[n+1] - \delta[n-4] \right) \\
&amp; = (\frac{1}{4})^{-n} \delta[n+1] - (\frac{1}{4})^{-n} \delta[n-4] \\
&amp; = (\frac{1}{4})^{-(-1)} \delta[n+1] - (\frac{1}{4})^{-(4)} \delta[n-4] \\
x[n] &amp; =  \frac{1}{4} \delta[n+1] - 256 ~ \delta[n-4] \\
\end{align}

So, your input sequence is $x[n] = \frac{1}{4} \delta[n+1] - 256 ~\delta[n-4] $

Convolving this with the impulse response of 
$$h[n] = \delta[n] - \delta[n-5]$$ is easy, this time by utilizing the shifting property of the impulse which is:

$$ f[n] \star \delta[n-d] = f[n-d] $$

Where $\star$ is th convolution. Then apply distribution of convolution over addition to get your result as:

\begin{align}
y[n] &amp;= x[n] \star h[n] \\
&amp; = x[n] \star (\delta[n] - \delta[n-5] ) \\
&amp;= x[n] - x[n-5] \\
&amp;= (\frac{1}{4} \delta[n+1] - 256 ~\delta[n-4]) - (\frac{1}{4} \delta[n-4] - 256~ \delta[n-9]) \\
&amp;= \frac{1}{4} \delta[n+1] - 256~ \delta[n-4] - \frac{1}{4} \delta[n-4] + 256~ \delta[n-9] \\
y[n] &amp;= \frac{1}{4} \delta[n+1] - (256~ \delta[n-4] + \frac{1}{4}) \delta[n-4] + 256~ \delta[n-9] \\
\end{align}

I hope this helps.