Define the total repayment period $T$ in units of years. Then the payments $P_t$ in any single year $t$ are given by:
$$P_t=12P_1(1+\alpha)^{t/2}$$
where $P_1$ is the initial payment ($\$282$ in your example above), and $\alpha$ is the constant percentage by which the payments go up every 2 years.
If you don't like interpolating a yearly increase in payments (interpolting like this will alter the output slightly), then you can replace the exponent $t/2$ with $q$ where
$$q=\left\{\begin{matrix}
t & if\:\:t\:\:is\:\:odd \\
t-1 & if\:\:t\:\:is\:\:even
\end{matrix}\right.$$
The outstanding loan balance in any year $t$ is then the polynomial expansion:
$$B_t=B_0x^t-\sum_{k=1}^{t} P_kx^{(t-k+1)}$$
where $B_0$ is the initial loan balance before any payments and $x=(1+r)$, where $r$ is the interest rate.
The loan is completely paid off in the $T^{th}$ year:
$$B_T=0$$
and so
$$B_0x^T=\sum_{k=1}^{T} P_kx^{(T-k+1)}$$
Solve this equation for $\alpha$ (you could use excel's solver to do this) to get the percentage that the payments should go up every 2 years.
As for your constraints, just make sure $P_1>W/2$ and $P_T<3W$ where $W$ is the "limiting amount" ($\$560$ in your example). You can also input these constraints easily into excel's solver.
In your example of $T=10$, $r=0.035$, $B_0=\$51183$ and $P_1=\$282$, I get $\alpha=0.3006$ (i.e. $30.06\% $)
EDIT: Note that the polynomial equation $$B_0x^T=\sum_{k=1}^{T} P_kx^{(T-k+1)}$$
can be expressed as a geometric progression:
$$B_0x^T=x^{T+1}12P_1\sum_{k=1}^{T} Q^k \:\: ; \:\:\: Q=y^{1/2}x^{-1}, \:\: y=(1+\alpha)$$
and thus as a more convenient formula which might be easier to solve for $\alpha$:
$$B_0x^T=x^{T+1}12P_1 \frac{Q-Q^T}{1-Q}$$
For this to work you have to be okay with interpolating a yearly increase in the payment amount.
EDIT #2:
The question asks for a way to determine the initial and final payments ($P_1$ and $P_T$) given the information $T$, $B_0$, and $r$. However, these three given values do not determine a unique $P_1$ and $P_T$.
You can rest assured, at least, that if $P_1$ is greater than your minimum bound ($\$560/2$) then all $P_t$ will be greater than your minimum bound. Likewise, if $P_T$ is less than your max bound ($3\times \$560 $), then so will all $P_t$.
If your $P_1$ is right on the minimum bound and this results in a final payment $P_T$ which is greater than your max bound, then you've got a problem. You will have to adjust the parameters in order to keep the payments in bounds.
