Solutions

a) Suppose for contradiction that this were not so; that there exists $b$ such that $f(b)=l\ne 0$.

Since $f$ is continuous, there exists $\delta>0$ such that

$$|f(x)-l|<|l| \implies |f(x)|>0$$

for all $x\in(b-\delta,b+\delta)$. Since this is an open interval, there would be a number $a$ in it. But this contradicts the premise that $f(y)=0$ for all $y$ in $A$.

b) Let $h(x)=f(x)-g(x)$. Since $h$ is continuous and $h(x)=0$ for all numbers $x$ in a dense set $A$, by a) we have that $h(x)=0$ for all $x$. But this means that $f(x)-g(x)=0$ and therefore $f(x)=g(x)$ for all $x$.

c) Suppose for contradiction that there exists $b$ such that $f(b)<g(b)$. We have $\lim_{x\to b}f(x)=f(b)$ and $\lim_{x\to b}g(x)=g(b)$. These limits are defined since $f$ and $g$ are continuous. Moreover, there exist $\delta_1,\delta_2>0$ such that

$$|f(x)-f(b)|<\frac{g(b)-f(b)}{2}$$

for all $x\in(b-\delta_1,b+\delta_1)$ and

$$|g(x)-g(b)|<\frac{g(b)-f(b)}{2}$$

for all $x\in(b-\delta_2,b+\delta_2)$.

Let $\delta=\min(\delta_1,\delta_2)$. Combining both inequalities, we have that $f(x)<g(x)$ for all $x\in(b-\delta,b+\delta)$. Since this is an open interval, there exists $y\in A$ with $y\in(b-\delta,b+\delta)$. But this contradicts the premise that $f(x)\ge g(x)$ for all $x\in A$.

We cannot replace $\ge$ with $>$. As a counterexample, consider $f(x)=x$ and $g(x)=0$ on the dense set $A=\{x:x\in\mathbb{R},\ x\ne 0\}$. Obviously, $x>0$ for all $x\in A$ with $x>0$, but not for $x=0$.