Solution Question 1: Let $f(1)-g(1)=\epsilon>0$ . Then, since $f$ is continuous, there exist a $\delta>0$ such that if $x\in(1-\delta,1]$ then

f(x)>f(1)-\epsilon=g(1)\geq g(x),

where we also used that $g$ is increasing in the last inequality.
Let us define the set

S=\{x\in[-1,1]\text{; s.t. }f(y)\geq g(y),\text{for }y\in[x,1]\}.

By the previous paragraph $(1-\delta,1]\subset S$ , thus $S\neq\emptyset$ , and by definition $S$ is bounded from below. Using the greatest lower bound property of the real numbers we may conclude that $x_{0}=glb(S)$ exists.
Next we note that $x_{0}>-1$ . This follows as in the first paragraph of the proof: by continuity of $f$ if $f(-1)+\hat{\epsilon}=g(-1)$ then there exist a $\hat{\delta}$ such that $f(x)>g(x)$ for all $x\in[-1,-1+\hat{\delta})$ and therefore $[-1,-1+\hat{\delta})\not\subset S$ . We can conclude that $x_{0}\in[-1+\hat{\delta},1-\delta]$ for some $\delta,\hat{\delta}>0$ .
To finish the proof we show that $f(x_{0})=g(x_{0})$ , that is $x_{0}$ solves the desired equation. First we take any sequence $x_{j}\in S$ s.t. $x_{j}\to x_{0}$ and make the following estimate

0\leq f(x_{j})-g(x_{j})\leq f(x_{j})-g(x_{0})\to f(x_{0})-g(x_{0}),

(1)

where we first used that $x_{j}\in S$ , then that $g$ is increasing and finally that $x_{j}\to x_{0}$ together with continuity of $f$ .
Similarly we notice that for each $j\in\mathbb{N}$ there is an $x_{j}$ such that $x_{0}-\frac{1}{j}\leq x_{j}\leq x_{0}$ and $f(x_{j})<g(x_{j})$ , since $x_{0}$ was the greatest lower bound of $S$ . Passing to the limit $j\to\infty$ we may conclude that

0>f(x_{j})-g(x_{j})\geq f(x_{j})-g(x_{0})\to f(x_{0})-g(x_{0}),

(2)

where we again used that $g$ is increasing and $f$ continuous. We can conclude from (2) that $f(x_{0})\leq g(x_{0})$ and from (1) that $g(x_{0})\leq f(x_{0})$ . It follows that $f(x_{0})=g(x_{0})$ .