From $f(g(2))=f(4)=11$, we have $4a+b=11$. Also $(g\circ f)(2)=(f(2))^2=49$ and $f(2)>0$ imply $2a+b=7$, so subtracting gives $a=2$; other options follow from taking the negative root or arithmetic slips.

Compute $f(g(x)) = \dfrac{2\cdot \frac{x}{x-2} - 1}{\frac{x}{x-2} + 3} = \dfrac{\frac{x+2}{x-2}}{\frac{4x-6}{x-2}} = \dfrac{x+2}{4x-6}$. Other choices drop the common factor, miss the factor 2, or miscompute the numerator.

Compute $g(5)=22$ and then $f(22)=\sqrt{22+4}=\sqrt{26}$. Reversing gives $g(f(5))=6$, and the others come from omitting the $+4$ or the square root.

First find $f(-1)=1+5=6$, then $g(6)=12+3=15$. The other values come from reversing to $f(g(-1))=-4$, forgetting the $+3$ to get 12, or stopping at $f(-1)=6$.

Substitute $g(x)$ into $f$: $f(g(x)) = 3\cdot \dfrac{1}{x} - 2 = \dfrac{3}{x} - 2$. The others come from reversing the composition or algebraic/sign errors.

Compute $g(3)=5$, then $f(5)=11$. Other choices come from reversing to $g(f(3))=45$, omitting the $+1$ to get 10, or misapplying operations to get 6.

Solve $y=\frac{2x-5}{x+3}$ for $x$ to get $x=\frac{3y+5}{2-y}$, hence $g(y)=\frac{3y+5}{2-y}$. The other choices either reuse $f$ (not an inverse) or have sign/placement errors in numerator or denominator.

We need $f(2x-3)=x^2-1$; let $u=2x-3\Rightarrow x=\frac{u+3}{2}$, so $f(u)=\left(\frac{u+3}{2}\right)^2-1=\frac{u^2+6u+5}{4}$ and $f(x)=\frac{x^2+6x+5}{4}$. The other choices ignore the inner shift, divide only by 2, or mishandle the expansion.

Compute $f(3)=1$ and then $g(1)=2$. The value $1/8$ comes from reversing to $f(g(3))$, and $10$ or $3$ come from using $g(3)$ or the original input instead of the composed output.

Since $f(x-1)=2x+3$, set $u=x-1$ to get $f(u)=2(u+1)+3=2u+5$, so $f(x)=2x+5$. The others come from reversing the composition ($2x+4$) or failing to account for the input shift ($2x+3$, $2x+1$).