Execuse the messy wiring!

$$O_0 = I_0 \oplus I_1 \oplus I_2 … \oplus I_7$$ $$O_1 =$$ left as an exercise to the reader $$O_3 =$$ left as an exercise to the reader

Toggles every negative edge as long as X is 1

This is left as an exercise to the reader.

$$A(t+1) = B \oplus C$$ $$B(t+1) = A$$ $$C(t+1) = C$$ $$ABC(0) = 100$$ $$ABC(1) = 010$$ $$ABC(2) = 101$$ $$ABC(3) = 110$$ $$ABC(4) = 111$$ $$ABC(5) = 011$$ $$ABC(6) = 001$$ $$ABC(7) = 100$$

Easy Solution: Convert the T-FlipFlop back to a D-FlipFlop! (i.e, implement a D-FlipFlop with a T-Flipflop)
Standard Solution: do the truth table, K-maps, equations, etc.
This is left as an exercise to the reader.

We can see that after 4 cycles, $A(t+4)=A + B + C$
We can also see the $C=$ carry of A’s sum
Initially $$A (0) = 0$$ $$C (0) = 0$$ After 4 cycles: $$A (4) = 1011 + 1011 + 0 = 0110$$ $$C (4) = 1$$

After 8 cycles: $$A (8) = 0110 + 1011 + 1 = 0010$$ $$C (8) = 1$$

After 12 cycles: $$A (12) = 0010 + 1011 + 1 = 0010$$ $$C (12) = 1$$

After 16 cycles: $$A (16) = 0110 + 1011 + 1 = 1110$$ $$C (16) = 0$$

Do the K-maps, then you get:
$$A = \bar{z} + x \oplus y$$ $$B = \overline{x \oplus y \oplus z}$$ $$C = x \oplus z$$ $$D = z + x \oplus y$$

$B$ can be simplifed as $$B = \overline{y \oplus C }$$ Which can be furthur simplified as $$B = \bar{y} \oplus C$$

Also if you are wondering how tf $\oplus$ and $\bar{\oplus}$ are implemented using AND and OR gates, refer to the book.

This is left as exercise to the reader.