$\text {(a)} $ Since $x_0^{2017}=1+x_0>1$, hence $x_0>1$. Also $y_0^{4034}=y_0+3x_0>3x_0>3\implies y_0>1$. We also have $x_0^{4034}=(1+x_0)^2=x_0+(1+x_0+x_0^2)$ so $x_0^{4034}-x_0=3x_0+(x_0-1)^2>3x_0=y_0^{4034}-y_0$. Put $f (t)=t^{4034}-t\implies f (x_0)>f (y_0) $. Now $f'(t)=4034t^{4033}-1>0\forall t\in\left [1,\infty\right) $, hence $f $ is increasing in $\left [1,\infty\right) $. But $x_0,y_0\in\left [1,\infty\right) $ and $f (x_0)>f (y_0)\implies x_0>y_0$ by increasing nature of $f $ in $\left [1,\infty\right) $. $\text {(b)} $ This is a good problem but still quite weak. We will show that $(x_0-y_0)<\frac {1}{10^{10}}\implies$ the required digit is $0$. From the first part we have $x_0^{4034}-x_0=3x_0+(x_0-1)^2=(y_0^{4034}-y_0)+(x_0-1)^2$. Using $x^k-y^k=(x-y)(\sum_{i=0}^{k-1}x^iy^{k-1-i})$, we see that $(x_0-y_0)(\sum_{i=1}^{4033}x_0^iy_0^{4033-i}-1)=(x_0-1)^2$. Call this result $(\mathcal {\Gamma}) $. Now put $(x _0-1)=a<1\implies (1+a)^{2017}=2+a$. Using $\text {Binomial Theorem}$, $(1+t)^n>1+nt\forall 0 <t<1,n\in\mathbb {Z}_+$, we obtain $(2+a)=(1+a)^{2017}>1+2017a\implies a <\frac {1}{2016} $. Also we have $\sum_{i=0}^{4033}x_0^iy_0^{4033-i}>\sum_{i=0}^{4033}y_0^{4033}=4034y_0^{4033}$ using $y_0 <x_0$. Also $y_0^{4033}=\frac {y_0^{4034}}{y_0}=\frac {y_0+3x_0}{y_0}>4$ using $x_0>y_0$. Thus $\sum_{i=0}^{4033}x_0^iy_0^{4033-i}>4034\cdot 4=16136$. Using the obtained results in $(\mathcal {\Gamma}) $ we obtain $(x_0-y_0)(16136)<\frac {1}{2016^2}\implies (x_0-y_0)<\frac {1}{10^{10}} $ using $16136>16000, 2016>2000,64>10$ so we are done and the answer is $0$. Using the same method for any general $n $ instead of $2017$ gives the bound $(x_0-y_0)<\frac {1}{8n^3} $.

Sorry, i rewrited problem.

Official solution for part b) We will use $n=2017$ $f(x)=x^{n}-x$ increase for $x>1$ $f(1+\frac{1}{n})=(1+\frac{1}{n})^n-1-\frac{1}{n}=1+1+ \frac{n(n-1)}{2n^2}+C_n-1-\frac{1}{n}=1+\frac{n-3}{2n}+C_n>1$ where $C_n$ some positive expression. $f(x_0)=0$ so $x_0<1+\frac{1}{n}$ $x_0^{2n}-x_0-(y_0^{2n}-y_0)=(1-x_0)^2<\frac{1}{n^2}$ $x_0^{2n}-y_0^{2n}=(x_0-y_0)(x^{2n-1}+...+y_0^{2n-1})>2n(x_0-y_0)$ $(2n-1)(x_0-y_0)\leq x_0^{2n}-x_0-(y_0^{2n}-y_0) < \frac{1}{n^2}$ So $x_0-y_0< \frac{1}{n^2(2n-1)}$ And for $n=2017$: $2017^2(2*2017-1)>16*10^9>10^{10}$