Solve for x and k. Model 1: log2(x) + 4 logx(8) = 8. Model 2: 3^(2k+1) - 28 3^k + 9 = 0

Model 1: $\log_2 x + 4 \log_x 8 = 8$
First, change the base of $\log_x 8$ to base 2 using the change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$:

$$\log_x 8 = \frac{\log_2 8}{\log_2 x} = \frac{3}{\log_2 x}$$

Substitute this into the equation:

$$\log_2 x + 4 \left(\frac{3}{\log_2 x}\right) = 8$$ $$\log_2 x + \frac{12}{\log_2 x} = 8$$

Let $u = \log_2 x$. The equation becomes:

$$u + \frac{12}{u} = 8$$

Multiply by $u$:

$$u^2 + 12 = 8u$$

Rearrange into a quadratic equation:

$$u^2 - 8u + 12 = 0$$

Factor the quadratic equation:

$$(u-2)(u-6) = 0$$

This gives two possible values for $u$: $u=2$ or $u=6$.
Substitute back $u = \log_2 x$:
If $u=2$, then $\log_2 x = 2 \implies x = 2^2 = 4$.
If $u=6$, then $\log_2 x = 6 \implies x = 2^6 = 64$.
The larger value of $x$ is 64.

Model 2: $3^{2k+1} - 28(3^k) + 9 = 0$
Rewrite $3^{2k+1}$ as $3^{2k} \cdot 3^1 = 3 \cdot (3^k)^2$:

$$3 \cdot (3^k)^2 - 28(3^k) + 9 = 0$$

Let $v = 3^k$. The equation becomes a quadratic equation in terms of $v$:

$$3v^2 - 28v + 9 = 0$$

Factor the quadratic equation:

$$(3v-1)(v-9) = 0$$

This gives two possible values for $v$: $v=\frac{1}{3}$ or $v=9$.
Substitute back $v = 3^k$:
If $v=\frac{1}{3}$, then $3^k = \frac{1}{3} \implies 3^k = 3^{-1} \implies k=-1$.
If $v=9$, then $3^k = 9 \implies 3^k = 3^2 \implies k=2$.
The problem states to use the larger value of $k$, so $k=2$.
Now, calculate the final thrust control $T$ using $T = 200k$:

$$T = 200 \cdot 2$$ $$T = 400$$

The thrust coefficient $k$ is 2, and the final thrust control $T$ is 400.

Model 3: $\sqrt{y+5} - \sqrt{y-2} = 1$
Isolate one square root term:

$$\sqrt{y+5} = 1 + \sqrt{y-2}$$

Square both sides of the equation:

$$(\sqrt{y+5})^2 = (1 + \sqrt{y-2})^2$$ $$y+5 = 1 + 2\sqrt{y-2} + (y-2)$$ $$y+5 = y - 1 + 2\sqrt{y-2}$$

Subtract $y$ from both sides:

$$5 = -1 + 2\sqrt{y-2}$$

Add 1 to both sides:

$$6 = 2\sqrt{y-2}$$

Divide by 2:

$$3 = \sqrt{y-2}$$

Square both sides again:

$$3^2 = (\sqrt{y-2})^2$$ $$9 = y-2$$ $$y = 9+2$$ $$y = 11$$

Check the solution: $\sqrt{11+5} - \sqrt{11-2} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1$. The solution is valid.
The value of $y$ is 11.

First, calculate the sum $x+k+y$:

$$x+k+y = 64 + 2 + 11 = 77$$

Now substitute this sum into the given relationship:

$$\log_2(77 - 13) = 6$$ $$\log_2(64) = 6$$

Since $2^6 = 64$, the statement $\log_2(64) = 6$ is true.
Thus, the sum of the results $(x+k+y)$ satisfies the given relationship.

The final answers are:

a) (i) The larger value of the fuel flow rate $x$ is $\boxed{64}$.

a) (ii) The thrust coefficient $k$ is $\boxed{2}$ and the final thrust control $T$ is $\boxed{400}$.

a) (iii) The value of $y$ is $\boxed{11}$.

b) The sum $x+k+y = 64+2+11 = 77$. Substituting into the relationship: $\log_2(77-13) = \log_2(64) = 6$. This shows the relationship is satisfied.