It is impossible.
Let the $7\times 28$ area be painted with black and white squares in a checkerboard pattern. Every piece will cover $2$ black and $2$ white squares, except the T-piece, which covers $3$ of one color and $1$ of another. Since there are $7$ T-pieces, a tiling that uses every piece cannot cover the same number of black and white ...
TLDR: I'll fill the board and prove that the solution is unique.
First, let's start by:
I'll paint those green:
Let's repeat those steps a few more times, using orange, blue, red and purple, in precisely that order:
We can easily fill the topmost white squares by that reasoning. They can't be filled in any other way:
Now, let's look at:
And by ...
This works! (I think)
(Hopefully that’s clear enough how the shapes go)
I got this mostly by thinking about how the blue and red can be placed such that the top and bottom of the green can be transported up and down without overlapping.
Brilliant idea! Hope to see some more!
The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this
or else you get this
The latter can be easily improved by replacing the ones at the edges to give this
A different way to get the same ...
I started with this:
Pushed things this way and that, ended up with this:
Similarly, on 9x9:
And on 10x10:
It took me a while to get there, but that one suggests an emerging pattern.
And here is an expandable solution for any $2n\times 2n$ grid.
FINITE PORTION OF ANSWER
This can be extended infinitely in all directions - see my route to solving below for how.
First a detour to explain how I made a tool (which competing answers could also use, perhaps improving on the finite number of differently-coloured tiles) by which ideas can be quickly tried using an Excel spreadsheet:
Select all, and set ...
Challenge 1: Fit the four yellow birds into the tray, no overlapping. Rotation and reflection are allowed.
Challenge 2: Fit the four yellow birds and the blue piece into the tray, no overlapping. Rotation and reflection are allowed.
Challenge 3: Fit the four yellow birds and the red piece into the tray, no overlapping. Rotation and reflection are allowed.
Here is a proof that $12$ is the smallest possible number of regions in any feasible solution.
Consider an arbitrary division of an arbitrary rectangle into $n$ regions, such that every region has exactly five neighboring regions. We translate this picture into a so-called planar graph: each of the $n$ regions then translates into a vertex/point, and there ...
Finally arrived at this solution after playing around on
for way longer than I care to admit!!!! :)
First pair (UI-TF)
Second pair (WX-PY)
Third pair (VZ-LN)
First, a generalizable solution for $1 \times n$, $n$ is even. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no W-pentominos can be shortened.
This is a way to tile
This is optimal for this $a \times b$ because
And here is a way to tile
I've found two more, one for 1x4:
and a rather large one ...
Some initial deductions:
In the bottom right,
And now, there's not much progress that can be made without thinking more globally.
Continuing with this newfound knowledge,
Finishing it off:
The final answer: