How are two signals multiplied? And how is it different from convolving two signals?
Yes, you are correct. Multiplication in time domain means convolution in frequency domain and vice versa.  Multiplying your signals $x[n]$ and $y[n]$ will give an output:

\begin{align}
z[n]&amp;=\{2\cdot 5, 4\cdot 1, 1\cdot 8\}\\ &amp;= \{10, 4, 8\}\end{align} 

Remember that this output is in time domain. When you convolve $x[n]$ and $y[n]$, you will get $z[n]$ in time domain as:

\begin{align}
z[n]&amp;=\{5\cdot 2, 5\cdot 4+1\cdot 2, 5\cdot 1+1\cdot 4+8\cdot 2, 1\cdot 1+8\cdot 4, 8\cdot 1\}\\ &amp;= \{10,22,25,33,8\}
\end{align}

Just flip one of the signals around zero and start moving right one place at a time. Multiply the corresponding points as you go along. The output has a larger sequence because convolution output has ${\rm length}(x)+{\rm length}(y)-1$ points.  

To answer your query about where to use multiplication &amp; convolution, assume you want to pass signal $x(n)$ through filter $y(n)$. The output of the filter $z(n)$ will be convolution of $x(n)$ and $y(n)$.
Now assume that you first converted from time to frequency domain, i.e. $X(e^{i\omega})$ and $Y(e^{i\omega})$ are frequency domain representation of $x(n)$ and $y(n)$. Now, to find $Z(e^{i\omega})$(output in frequency domain), you have to multiply $X(e^{i\omega})$ and $Y(e^{i\omega})$. To get the output in time domain i.e. $z(n)$ you have to apply inverse transform.  

So that's how you use convolution and multiplication.