Calculating a delayed response of a system
The answer @Juancho liked to is indeed very useful. To me it looks like you don't understand time invariance fully, and that answer sould clarify things.

But, concerning your questions of getting the delayed response of the systems:

With the first system, $y[n] = x[n] + x[n-1] + x[0]$, you are right. The constant term will not change, since it is constantly $x[0]$, and thus not dependent of $n$. The correct delayed response is:

$$y[n-1] = x[n-1] + x[n-2] + x[0]$$

You can calculate that by substituting $n$ by $m-1$ (or $n-1$, it's the same which letter you use, but it's simpler for me to use a different letter while substituting). Now, whit $n=m-1$, you have:

$$y[n] = y[m-1] = x[(m-1)] + x[(m-1)-1] + x[0] = x[m-1] + x[m-2] + x[0]$$

When you get the final result (when you can't simplify any more), you can change the variable name from $m$ to $n$ again and get the first formula above.



With the second system, you employ the same logic to get the delayed response. Starting from:

$$y[n] = x[2n]$$

To get a delayed response you must substitute $n$ with $m-1$ and you get:

$$n = m-1$$
$$y[n] = y[m-1] = x[2 \times (m-1)] = x[2m - 2]$$

and now you can change the variable name again to get:

$$y[n-1] = x[2n - 2]$$.

These are explanations on how to calculate a delayed response (by 1). Looking at a response of a system delayed by 1 actually means that at time $n=5$, you will be looking at a response that you would normally get for $n=4$. The system is delayed, i.e. it behaves as the original, just with some time-delay.

To put it very simply, if I mark the delayed response by $y'[n]$, and the original by $y[n]$, this would hold:

$$y'[1] = y[0]$$
$$y'[2] = y[1]$$
$$y'[100] = y[99]$$

and more generally:

$$y'[n] = y[n-1]$$

For understanding time-invariance in systems, consult the linked question since that's precisely its topic.



To comment on the incorrect answers you proposed as delayed responses: they are actually responses to a delayed input.

What that means is that you will be feeding your original system with a delayed input. If your original input signal was $x[n]$, to get a response to a delayed input, you actually want to "feed" your system to a signal $x'[n] = x[n-1]$. Let's mark systems response to a delayed input with $y''[n]$ and calculate that for your first example:

$$ y''[n] = x'[n] + x'[n-1] + x'[0]$$

The system didn't change: It's still calculating the same operation: summing up the input at time $n$ with the input at time $n-1$ with the value of the input at time $0$. But, the input changed!. Substituting $x'[n] = x[n-1]$, you get:

$$ y''[n] = x[(n)-1] + x[(n-1)-1] + x[(0)-1] = x[n-1] + x[n-2] + x[-1]$$



To summarize, in order for a system to be time-invariant, the response of the system delayed by $k$ should be the same as the response of the system to the input delayed by $k$.

In this example, we were just examining delays by $1$. What proves that the system is not time-invariant in you case is the fact that $y'[n] \neq y''[n]$ because you have:

$$y'[n] = y[n-1] = x[n-1] + x[n-2] + x[0]$$
$$y''[n] = x'[n] + x'[n-1] + x'[0] = x[n-1] + x[n-2] + x[-1]$$