1. Let $t=x-1$ so that $x\to 1$ implies $t\to 0$.

Then the base becomes $\frac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}$ Substitute $x=1+t$: $-ax+\sin(x-1)+a=-a(1+t)+\sin t+a=\sin t-at$ and $x+\sin(x-1)-1=(1+t)+\sin t-1=t+\sin t$ Hence the base is $B(t)=\frac{\sin t-at}{t+\sin t}.$

Also, the exponent is $\frac{1-x}{1-\sqrt{x}}=\frac{-(x-1)}{1-\sqrt{x}}.$ Using $1-x=(1-\sqrt{x})(1+\sqrt{x}),$ we get $\frac{1-x}{1-\sqrt{x}}=1+\sqrt{x}\to 2 \qquad (x\to 1).$ So the limit is of the form $\lim_{x\to 1} \big(B(x)\big)^{1+\sqrt{x}}.$ If $B(x)\to L>0$, then the limit equals $L^2$.

2. Now compute the limit of the base. As $t\to 0$, use $\sin t\sim t$: $\sin t-at \sim (1-a)t,$ $t+\sin t\sim 2t.$ Therefore $\lim_{t\to 0} B(t)=\frac{1-a}{2}.$

Thus, provided this limit is positive, the whole expression tends to $\left(\frac{1-a}{2}\right)^2.$ We need $\left(\frac{1-a}{2}\right)^2=\frac14.$ So $\frac{1-a}{2}=\pm \frac12.$ Hence $1-a=\pm 1.$ This gives:

• $1-a=1 \Rightarrow a=0$,
• $1-a=-1 \Rightarrow a=2$.

3. Check validity of both values. The expression involves a real power with exponent tending to $2$ but not exactly constant integer, so for the real-valued limit near $x=1$, the base must be positive in a neighborhood of $x=1$.

• For $a=0$: $B(t)=\frac{\sin t}{t+\sin t}\to \frac12>0,$ so the limit is $\left(\frac12\right)^2=\frac14.$ Valid.

• For $a=2$: $B(t)=\frac{\sin t-2t}{t+\sin t}\to -\frac12<0.$ Near $x=1$, the base is negative, while the exponent $\frac{1-x}{1-\sqrt{x}}=1+\sqrt{x}$ is generally non-integer. Then $B(x)^{1+\sqrt{x}}$ is not real-defined in a neighborhood of $x=1$, so the given real limit does not exist. Thus $a=2$ is not admissible.

4. Therefore the only non-negative integer satisfying the condition is $a=0.$ Hence the largest such value is $\boxed{0}.$