Okay - I get now that you talking about changing the ground. But still this princicple applies: the same transformation that you applied to the ground is the transformation that applies to the ball.
You rotated the plane. I don't know how you have accomplished that rotation in your program, but mathematically it is expressed as a matrix $M$ that satisfies $M^TM = I$, the identity matrix (where $M^T$ is the transpose of $M$). If $\bf n$ is the normal vector to the original plane, then $M\bf n$ is the normal vector to the rotated plane.
You also said you moved the plane (in the example, along the $x$-axis). This means that the new plane no longer passes through the origin (at least in your example, the original point did). This is a translation, but we need to be careful which translation we use. Suppose the new plane must pass through some point $\bf p$. In your example you indicated that it passes through $(\sim 0.5, 0, 0)$. You can use that to be the point $\bf p$. Since the normal vector is $M\bf n$ and it passed through $\bf p$, a point $\bf r$ is on the new plane if $$(M{\bf n})^T{\bf r} = {\bf n}^TM^T{\bf r} = d$$ where $d = {\bf n}^TM^T\bf p$. The translation vector we need is ${\bf b} = dM\bf n$, which is the closest point on the new plane to the origin.
So the tranformation to the original plane ${\bf n}^T{\bf r} = 0$ to get the new plane ${\bf n}^TM^T{\bf r} = d$ is to rotate by $M$, then add the translation vector $b$:
$$ {\bf r} \mapsto M{\bf r} + \bf b$$
(In case you don't know, $\mapsto$ means "maps to". I.e., the original value on the left is transformed into the value on the right.) This is the same transformation you need to apply to your sphere to get it's new location after the transformation. If $\bf c$ is the original center of the sphere, the transformed center will be $${\bf c} \mapsto M{\bf c} + \bf b$$
Let $\bf v$ be the "input direction vector" of the horizontal ball movement. Since $\bf v$ is a direction, not a point, it does not translate. So the new direction vector that the ball will travel up the sloped ground will be given by $${\bf v} \mapsto M\bf v$$
So the inputs you need to solve your problem are the rotation vector $M$, and a point $\bf p$ that the new plane passes through. If you have these, then you can use them to transform the ground plane, the sphere's position, and the direction of movement, all three.
I think you have the forces of the stream and the wind correct (although it's hard to believe they want you to give them equal weight, but whatever ...) The problem is with the rowing.
The rower will row with some velocity $a\vec{i}+b\vec{j}$ He wants to row straight across the stream, so
$v_i + w_i + a_i = 0$.
We still need a condition on $b$. We are told that he can row 5 km/hr. Therefore
$a^2 + b^2 = 25$.
With those two equations, you should be able to solve the problem.
Best Answer
Remember that the center of mass (assuming all trees have the same mass) is $\frac{1}{5}(A + B + C + D + E)$ (treating the locations as vectors).
After the first step, we end up with the point $\frac{1}{2}(A + B)$, found by taking their midpoint. Now from here, we travel $\frac{1}{3}$ of the distance to $C$. We take the weighted average to get the new position of $$\frac{2}{3}\left(\frac{1}{2}(A + B)\right) + \frac{1}{3}C = \frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C$$ Similarly, after the next step, we end up at $$\frac{3}{4}\left(\frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C\right) + \frac{1}{4}D = \frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D$$ Likewise, the final step brings us to $$\frac{4}{5}\left(\frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D\right) + \frac{1}{5}E = \frac{1}{5}A + \frac{1}{5}B + \frac{1}{5}C + \frac{1}{5}D + \frac{1}{5}E$$ Intuitively, each of the vectors adds less proportional weight each time, while those already used have their contributions slightly diminished each time. This has the effect of balancing out in the end to give the average.