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] &= (\frac{1}{4})^{-n} \left(\delta[n+1] - \delta[n-4] \right) \\
& = (\frac{1}{4})^{-n} \delta[n+1] - (\frac{1}{4})^{-n} \delta[n-4] \\
& = (\frac{1}{4})^{-(-1)} \delta[n+1] - (\frac{1}{4})^{-(4)} \delta[n-4] \\
x[n] & = \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] &= x[n] \star h[n] \\
& = x[n] \star (\delta[n] - \delta[n-5] ) \\
&= x[n] - x[n-5] \\
&= (\frac{1}{4} \delta[n+1] - 256 ~\delta[n-4]) - (\frac{1}{4} \delta[n-4] - 256~ \delta[n-9]) \\
&= \frac{1}{4} \delta[n+1] - 256~ \delta[n-4] - \frac{1}{4} \delta[n-4] + 256~ \delta[n-9] \\
y[n] &= \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.
