The transfer function and impulse response for system A seem to be correct; good that you also noticed the pole zero cancellation. You could manually calculate the impulse response from the transfer function using the inverse Laplace transform, but in practice it is often easier/quicker to use a Laplace transform table such as this one. From such a table you should also be able to derive what the (minimal) transfer function should be for system B.
In order to answer question 1 you could indeed do the convolution of the two impulse responses, however I believe that one (me) can easily make mistakes when solving integrals. Instead I would recommend to decompose the total transfer function into a sum of simple terms that appear in a Laplace transform table. For example if the total transfer function is given by
$$
G(s) = \frac{a_2\,s^2 + a_1\,s + a_0}{(s + b_1)^2\,(s + b_2)},
$$
then you would want to solve for $A$, $B$ and $C$ such that
$$
G(s) = \frac{A}{s+b_1} + \frac{B}{(s+b_1)^2} + \frac{C}{s+b_2}.
$$
The total impulse response can then be found by summing the impulse responses of each of the three terms, where impulse response of each term should be able to be found on a a Laplace transform table.
Question 2 does not have a completely well defined answer, because no initial conditions are given for system A and B. One way you could interpret this is with all initial conditions set to zero, which can be obtained by evaluating the convolution integral between the input and the impulse response of the total system. Another way is to assume that the output has settled into a period signal as well, which can be obtained by first evaluating the Fourier series of the input
$$
x(t) = \sum_{k=0}^\infty a_k \cos(2\,\pi\,k\,t) + b_k \sin(2\,\pi\,k\,t) = \sum_{k=0}^\infty c_k \cos(2\,\pi\,k\,t + \phi_k),
$$
the output can then be obtained using
$$
y(t) = \sum_{k=0}^\infty c_k\,|G(j\,2\,\pi\,k)| \cos(2\,\pi\,k\,t + \phi_k + \angle G(j\,2\,\pi\,k)),
$$
so the total transfer function is evaluated at $s=j\,2\,\pi\,k$, with $j$ the imaginary unit.
This second approach would basically already give the answer to question 3, so I guess the first approach, with the convolution integral, is probably the intended answer.
There is no problem with the computations above, but there is definitely a misunderstanding in the way they are interpreted.
- When one considers the response $\Psi(s)=H(s)X(s)$ of the system , you are essentially saying that the system is satisfying a differential equation with the unperturbed initial condition $y(0)=0$, in your case the equation is $y'(t)+2y(t)=x(t)~,~ y(0)=0$. Laplace transforming this equation gives the exact expression above. The solution you have found is correct and satisfies this problem, and you can check that explicitly.
- On the other hand saying that the solution of this problem is $x(t)/\lambda$ where $\lambda$ is the eigenvalue of the function $x(t)$ with respect to the linear operator $T=\frac{d}{dx}+2$ is NOT correct. What this function represents is a particular solution to the inhomogeneous ODE, and in no way a complete solution to the problem. In your case it can be shown that the full solution to the ODE mentioned above is
$$y(t)=Ce^{-2t}+\frac{e^{2jt}}{2j+2}\equiv y_0(t)+y_p(t)~~,~C\in\mathbb{R}$$
where $C$ is an arbitrary constant which is determined by the initial conditions. The only way $y_p(t)$ can be a solution by itself is if there is an initial condition that sets $C=0$. In this case no real number can achieve this, so the inhomogeneous solution will always be accompanied by some sort of term $\propto e^{-2t}$.
- As far as the convolution integrals are concerned, since all signals are right-sided one can show that the following three expressions are equivalent:
$$\Psi(s)=H(s)X(s)\iff y(t)=\int_0^{t}dt' h(t')x(t-t')\iff y'(t)+2y(t)=x(t)~,y(0)=0$$
which you have shown to be true. Exchanging the order of $h,x$ does not change anything.
However, the interpretation of the same convolution integral from $(0,\infty)$ is completely different. To obtain that integral you secretly assume that $x(t)$ is not right-sided, and instead has preexisted for all times $t\in \mathbb{R}$. This means that the system started getting kicked in the infinite past, which means that it must have reached the equilibrium solution at any finite time (all transients have died since the system is causal) and that is the reason why you obtain the equilibrium solution when you perform the integral in $(0, \infty)$ instead. Indeed you can demonstrate this using the ODE representation:
$$y'(t)+2y(t)=x(t)~,~ y(t_0)=y_0 \Rightarrow y(t)=e^{2t_0}\left(y_0-\frac{e^{2jt_0}}{2j+2}\right)e^{-2t}+\frac{e^{2jt}}{2j+2}$$
Taking the limit $t_0\to -\infty$ requires the transients to vanish
$$\lim_{t_0\to-\infty}y(t)=\frac{e^{2jt}}{2j+2}$$
and only the particular solution remains. It is not fair to ask both convolution integrals to give you the same answer, because they are designed to answer different problems. The method of particular solutions-eigenfunctions yields the long-time behavior of a causal system, the Laplace transform is useful for studying the full response of a system.
- Finally, in the last example the method of eigenfunctions simply does not apply, because $e^{-2t}$ is a solution to the homogeneous ODE $y'+2y=0$, which doesn't allow the application of the method. Regardless of that subtlety, if we follow the same procedure we used above, we see that in this problem the initial condition $y(0)=0$ sets the arbitrary constant to zero, and thus renders the two solutions equal, but this is a coincidence owing to the fact that the particular solution $y_p(t)=t e^{-2t}$ has $y_p(0)=0$.
Best Answer
There is a way to do this analytically using Fourier transforms. I will show this as soon as I can, but I promise you, it will be a lot messier than what I present below.
GRAPHICAL METHOD
To do this at the level of an undergrad signals class, you really need to draw a picture. But while it may be error-prone, there are checks along the way to see that you are doing it correctly.
The convolution integral is given by
$$x*h(t) = \int_{-\infty}^{\infty} dt' \, x(t-t') h(t') $$
The key here is to figure out what $x(t-t')$ is. I will let you figure out why
$$x(t-t') = \begin{cases} 1 & t-T \lt t' \lt t \\ 0& (t' \lt t-T) \cup(t'\gt t) \end{cases} $$
Now we may resolve the integral graphically as the area of intersection of the rectangle $x$ and the triangle $h$. You should see that $x*h(t)=0$ when $t \lt 0$. Next, note that, as we increase $t$, the region of intersection is a right triangle and the convolution integral is
$$\int_0^t dt' \, 1 \cdot t' = 1/2 t^2$$
This is true until $t=T$, when the left side of the rectangle crosses the axis. When $t \gt T$, the integral is now
$$\int_{t-T}^t dt' \, 1 \cdot t' = T \left (t-\frac{T}{2} \right )$$
When $t \gt 2 T$, the right side of the rectangle extends out past the triangle, so the convolution integral is now
$$\int_{t-T}^{2 T} dt' \, 1 \cdot t' = \frac12 (3 T^2+2 T t-t^2)$$
When $t \gt 3T$, the convolution is zero.
Please observe this by drawing the graphs. It may take several drawings, but you will see this.
To sum, we have
ANALYTICAL METHOD
Here we may use the convolution theorem for Fourier transforms:
$$\int_{-\infty}^{\infty} dt' \, h(t') x(t-t') = \frac1{2 \pi} \int_{-\infty}^{\infty} d\omega \, H(\omega) X(\omega) \, e^{-i \omega t} $$
where $H$ and $X$ are the respective Fourier transforms of $h$ and $x$. In this case:
$$X(\omega) = \int_0^T dt \, e^{i \omega t} = \frac{e^{i T \omega}-1}{i \omega} $$ $$H(\omega) = \int_0^{2 T} dt \, t \, e^{i \omega t} = \left (\frac1{\omega^2} - \frac{i 2 T}{\omega} \right ) e^{i 2 \omega T} - \frac1{\omega^2} $$
Thus, after some algebra, we find the convolution integral to be
$$\frac1{i 2 \pi} \int_{-\infty}^{\infty} d\omega \, \left [e^{i (3 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) - e^{i (2 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) - \frac1{\omega^3} e^{i (T-t) \omega} + \frac1{\omega^3} e^{-i t \omega} \right ] $$
This looks like a heck of an integral to do out. One way we can attack it is to use contour integration in the complex plane. (I warned you.)
So let's first consider
$$PV \int_{-\infty}^{\infty} d\omega \, e^{i (3 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) $$
Note that, when we break the integral apart, each piece on its own does not converge. So we consider the Cauchy principal value of the integral, knowing that when we put everything back together, the singular pieces should drop out and we will have our nice convolution integral again.
So now consider the contour integral
$$\oint_C dz \, e^{i (3 T-t) z} \left (\frac1{z^3} - \frac{i 2 T}{z^2} \right ) $$
where $C$ is a semicircular contour of radius $R$ in the upper half plane with a small semicircular detour of radius $\epsilon$ about the origin into the contour when $t \lt 3 T$ and $C$ is a semicircular contour of radius $R$ in the lower half plane with a small semicircular detour of radius $\epsilon$ about the origin into the contour when $t \gt 3 T$.
(If you need to understand why the contours need to be in the upper or lower half-planes like this, there are lots of resources on M.SE which address this very point. I will not do it here as it takes away from the discussion at hand.)
When $R \to \infty$ and for small $\epsilon$, we evaluate the contour integral in each case and apply Cauchy's theorem, i.e., the contour integral is zero. In this example, I will write everything out - here, let's use the upper contour:
$$PV \int_{-\infty}^{\infty} d\omega \, e^{i (3 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \, e^{i (3 T-t) \epsilon e^{i \phi}} \left (\frac1{\epsilon^3 e^{i 3 \phi}} - \frac{i 2 T}{\epsilon^2 e^{i 2 \phi}} \right ) =0$$
Now, we expand out the exponential, collect like terms in $\epsilon$, and integrate. The result is, for small epsilon and $t \lt 3T$
$$\frac1{i 2 \pi} PV \int_{-\infty}^{\infty} d\omega \, e^{i (3 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) + \frac1{\pi} (t-T) \frac1{\epsilon} +\frac14 (t^2-2 T t-3 T^2) = 0$$
For $t \gt 3 T$, we now integrate over the detour from $\pi$ to $2 \pi$ instead, and the result is
$$\frac1{i 2 \pi} PV \int_{-\infty}^{\infty} d\omega \, e^{i (3 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) + \frac1{\pi} (t-T) \frac1{\epsilon} -\frac14 (t^2-2 T t-3 T^2) = 0$$
OK, you may have also noticed that we have a term in $1/\epsilon$, which becomes singular as $\epsilon \to 0$. That's OK, because we have other pieces of the integral which will also have singular pieces, and we expect the singular parts to cancel prior to taking the limit as $\epsilon \to 0$.
The other integrals I will just state and allow the interested reader to work out for him/her/themself:
$$t \lt 2 T$$ $$-\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{i (2 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) - \frac{t}{\pi} \frac1{\epsilon} - \frac1{4} (t^2-4 T^2) = 0 $$
$$t \gt 2 T$$ $$\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{i (2 T-t) \omega} \left (\frac1{\omega^3} - \frac{i 2 T}{\omega^2} \right ) - \frac{t}{\pi} \frac1{\epsilon} + \frac1{4} (t^2-4 T^2) = 0 $$
$$$$
$$t \lt T$$ $$-\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{i (T-t) \omega} \frac1{\omega^3} - \frac{t-T}{\pi} \frac1{\epsilon} - \frac1{4} (T-t)^2 = 0 $$
$$t \gt T$$ $$\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{i (T-t) \omega} \frac1{\omega^3} - \frac{t-T}{\pi} \frac1{\epsilon} + \frac1{4} (T-t)^2 = 0 $$
$$$$
$$t \lt 0$$ $$\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{-i t \omega} \frac1{\omega^3} + \frac{t}{\pi} \frac1{\epsilon} + \frac1{4} t^2 = 0 $$
$$t \gt 0$$ $$\frac1{i 2 \pi}PV \int_{-\infty}^{\infty} d\omega \, e^{-i t \omega} \frac1{\omega^3} + \frac{t}{\pi} \frac1{\epsilon} - \frac1{4} t^2 = 0 $$
Now, we have to combine the above results to form the final convolution integral. It is worth laying out exactly how we are going to combine these results. I will illustrate with the following array:
$$\begin{array}\\ & 0 & T & 2 T & 3 T \\ t < 0 & < & < & < & < \\ 0< t < T & > & < & < & < \\ T \lt t \lt 2 T & > & > & < & < \\ 2 T \lt t \lt 3 T & > & > & > & < \\ t \gt 3 T & > & > & > & > \end{array} $$
The $\lt$ or $\gt$ denotes which of the results (e.g., $t \gt 3 T$ or $t \lt 3 T$) we are using for the value of $t$ in the respective row. Note that, no matter how we combine the results, the singular $1/\epsilon$ terms will cancel.
So, for example, for $t \lt 0$, we combine all of the "less than" results above. And, fortunately, we find that everything cancels and the result is zero, as we expect. In fact, I leave it as an exercise for the reader to verify that, by combining the results as specified in the above matrix, that we reproduce the results obtained by computing the integral graphically.
Still want to do the convolution integrals analytically?
ADDENDUM
Here's a plot for the case $T=2$: