PG,
When I look at your comments, I feel that there must be an underlying misconception that has generated this series of questions. I think our first task is to identify this misconception, and I'm going to take a guess at what it might be.
As I read what you wrote, I wonder if you are forgetting that there are two different meanings of integration.
First, there is the idea of an "anti-derivative". If we have a function like y=x^2, we know the derivative is dy/dx=2x. The anti-derivative of 2x is then x^2 + C. Notice that when we reverse the operation, we need to add a constant C because the derivative of a constant is zero, and so any value of C provides a valid anti-derivative. We also call the anti-derivative operation an "indefinite integral". When you look at a table of indefinite integrals, you might see the following, although sometimes the constant C is left out because it is known to always be present.
[latex]\int x\;dx=\frac{x^2}{2}+C[/latex]
Second, there is the idea of a "definite integral" and this is the idea of area under a curve. Here, we must specify two limits for the integration, and the constant C is not relevant because it will subtract out. Let's do a simple definite integral of y=x over the range of 0 to 1. Clearly, we know the answer is 1/2 just by looking at the graph, but let's do it formally and notice how any constant C gives the same answer because it subtracts out. Since is always subtracts out, it is traditional to use C=0 when doing definite integrals.
[latex]\int_0^1 x\; dx=\left[\frac{x^2}{2}+C \right] _0^1=(\frac{1^2}{2}+C)-(\frac{0^2}{2}+C)=\frac{1}{2}[/latex]
Now, how does this relate to your questions?
For question Q, notice that your question 14 says to integrate from negative infinity to some value of "t". This implies a definite integral over a range, and the area is the cumulative area under the curve from negative infinity to a particular point. Let's write this integral out formally and notice the use of a dummy integration variable tau.
[latex]h(t)=\int_{-\infty}^t g(\tau)\; d\tau[/latex]
Notice that even though the function is g(t), we put it in the integral as a function of tau and integrate over the variable tau. Here, tau is a dummy integration variable that goes away after the operation, and you are left with a new function of "t", which I called h(t) above.
The meaning of h(t) is the area under the curve g(t) from negative infinity up to the point "t". So, this is why the area can't be zero in your example when t>2.
Now, what about Q1? This is a bit of a silly question, but I realize you are just trying to rationalize how to get a particular result. We never make area negative just because t<0.
Now, for Q2. Hopefully, my above description of area being a definite integral answers this. You can't just say the area for the rect(t) function is "t" and therefore the area at t=0 is zero. No, you have to do out a full definite integral with proper limits to get area.
So, let's do it out. First, rect(t)= zero for t<1/2, so instead of starting the integration at negative infinity, we can start at -1/2. All area up to t=-1/2 is zero.
[latex]A=\int_{\frac{-1}{2}}^t {\mathrm {rect}}(\tau)\; d\tau=\left[t\right] _{\frac{-1}{2}}^t={t}+\frac{1}{2}[/latex] for -1/2<t<1/2, and A=1 for t>1/2 and A=0 for t<-1/2
Notice that I restrict this answer t+1/2 to be for -1/2<t<1/2 because the "rect" function goes to zero outside this range and the antiderivative is zero (strictly it is any constant number, but it subtracts out) there. Clearly, the area is one for t>1/2. You can see this visually in the plot, or you can do it formally with definite integrals.
