Very interesting question! I'll start by outlining some of the mathematical basics for this question:
You're looking for a bounded state of some potential $V$, i.e. a non-scattering state. Mathemematically, this translates to an $L^2$-integrable eigenfunction of the Schrödinger operator $-\Delta+V$.
By elliptic regularity, for these functions you instantly get what I understand as your condition (3) (The more precise statement would be that $\psi$ lies in the domain of the self-adjoint version of $\hat{p}$). Basically, the argument here is, that $\psi$ has to be two-times differentiable, by rearranging the Schrödinger equation, for your class of potentials you will have $\Delta \psi\in L^2$, by Fourier transform you then get that the first derivative will also be $L^2$. Hence, $\hat{p}$ is well-defined for $\psi$.
The basic idea why most bound states you will come across are exponentially decaying comes from the following idea: Assume that far away from the origin, $V$ is monotone i.e. it doesn't oscillate.
This allows us toestimate $V$ from below by a box potential, which implies that a bounded state of $V$ will be dominated by a bounded state of the box potential. Bounded states of box potentials do exponentially decay, hence the state will decay exponentially. This argument can be made explicit using maximum principles for elliptic PDEs, you may up the mathematical details in e.g.
Berezin and Shubin, The Schrödinger equation (Springer 1991).
So from this argumentation, the answer to your question is almost no for potentials which are monotone far outside. By "almost", I mean that there may be such functions at distinguished, but physically irrelevant values of $E$, for example, consider the potential
$$
V(x)=\frac{2-6x^2}{(1+x^2)^2}
$$which looks like this:
You may now check that $\psi(x)=\frac{2}{\sqrt{\pi}}\frac{1}{1+x^2}$ is a normalized eigenfunction to this potential with eigenvalue 0. The momentum operator is well-defined for this $\psi$ and $\psi$ obviously decays only polynomially for $x\rightarrow\infty$. So, what happened here? If you try to use the "box" argument, you would compare to a box which is completely negative away from the origin (remember, the box estimates the potential from below), so 0 is already a scattering state for the box! However, looking at the potential, you see that this can only be the case for this exact value of $E$ - for even an $\epsilon$ more energy, you will obtain a scattering state since $V\rightarrow 0$ for $x\rightarrow \infty$; and for an $\epsilon$ less, you will get a bound state you can again estimate by a box, hence it decays exponentially. Since you can't prepare a state with an exact energy, this is not physically relevant. Generally speaking, this phenomenon should only occur at $E=\limsup_{|x|\rightarrow\infty}V(x)$, since this will correspond to the lowest possible energy for scattering states.
So, what can happen if we drop the "monotonicity-far-outside"-condition? I think that in this case, it should be possible to obtain the kind of states you are searching for. My attempt on the construction goes as follows: Let $V$ be a collection of box potentials where the boxes have constant height and grow thinner as $x$ gets bigger, e.g. something that looks vaguely like this:
If the infinitely many discontinuities bother you, the behavior I'll describe should be exactly the same for a smoothed version of this potential.
Now, a bound state of this potential would oscillate around $0$ where $V=-0.5$ and decay where $V=+0.5$. The (exponential) decay rate where $V=0.5$ is always the same, by controlling the width of the boxes, you can exactly control how fast your bound state decays, e.g. you can achieve, that each time you pass the positive part of the box, your amplitude goes down at the rate $1/x^2$.
The details are probably very technical and fishy, but I think in principle this should work.
There is nothing to "derive" for the kinetic energy operator. By definition, classical kinetic energy is $\frac{p^2}{2m}$, and so $\hat{E}_\text{kin} = \frac{\hat{p}^2}{2m}$ quantumly. It's not exactly clear why you think this doesn't make sense mathematically, but it does: In words, it says "apply the momentum operator twice, then divide the result by $2m$".
Note that $\langle p^2\rangle \neq \langle p\rangle^2$, the difference is precisely what the standard deviation is defined as and what is usually called the "uncertainty" $\Delta p$ in most physics texts.
Best Answer
You can start by finding a suitable expression for the function.
$$ \psi(x)=\begin{cases} k & |x|\lt a \\ \frac{k}{b}(x+a+b) & -a\lt x\lt -(a+b) \\ \frac{k}{b}(a+b-x) & a\lt x\lt (a+b) \end{cases} $$
Then you need to normalize the wavefunction and find the value of $k$. Do it by integrating region by region, in a piecewise manner. You should get $k^2=\frac{3}{2(3a+b)}$
Now you need to find $\langle\psi|\frac{p_{x}^2}{2m}|\psi\rangle$
This is equivalent to :
$$\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx$$
From here, you have two options. You note that, using integration by parts :
$$\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx=\psi(x)\int\frac{\partial^2\psi(x)}{\partial x^2}dx|_{-\infty}^{\infty}\space -\int_{-\infty}^{\infty}\frac{\partial\psi(x)}{\partial x} (\int_{-\infty}^{\infty}\frac{\partial^2\psi(x)}{\partial x^2}dx)dx$$
The first term on the RHS goes to $0$, as $\psi(x)\rightarrow 0$ at both the infinities. Hence, you have :
$$\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx=\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}(\frac{\partial\psi(x)}{\partial x})^2dx$$
You can easily calculate $(\frac{\partial \psi(x)}{\partial x})^2$ in these intervals. After integrating, you should get $\frac{\hbar^2 k^2}{mb}$.
Plugging the value of $k$ that you got from normalization, you should get the desired expression.
Another alternative is to note that, the double derivative of a 'kink' is a Dirac delta function. Hence, you should have :
$$\frac{\partial^2\psi(x)}{\partial x^2}=-\frac{k}{b}(\delta(x+a)+\delta(x-a))$$
You can plug this into the integral, and get the same thing out, by noting $\psi(a)=\psi(-a)=k$
You can solve the integral in this way too, and reach the final expression.