Answer
(a) Please see below.
(b) Please see below.
Work Step by Step
(a)
By the assumption, $\lim_{x \to 0} (3x+1)(3x-1)x^2+0.01 =0.01$; that is, for any $\epsilon > 0$, there exists a $\delta >0$ such that$$|x-0|< \delta \quad \Rightarrow \quad |((3x+1)(3x-1)x^2+0.01)-(0.0.1)|< \epsilon .$$The last inequality implies that$$- \epsilon < ((3x+1)(3x-1)x^2+0.01)-(0.0.1) \quad \Rightarrow \quad 0.01- \epsilon < (3x+1)(3x-1)x^2+0.01.$$Now, by letting $\epsilon = \frac{0.01}{2}$, it follows that$$0.01-\frac{0.01}{2}=\frac{0.01}{2} < (3x+1)(3x-1)x^2+0.01.$$Since the limit exists, there exists some $\delta _{0}$ corrosponding to $\epsilon = \frac{0.01}{2}$. By the above inequality we conclude that for all $x \neq 0$ in the interval $(-\delta _0 , \delta _0)$, we have $(3x+1)(3x-1)x^2+0.01 > \frac{0.01}{2} >0$ (Please note that we excluded the point $x=0$ from this interval since in the definition of "limit" of a function $f$ at the point $c$, the function may not be defined at $c$ or may have any value, which does not have any relation to the existence of the limit).
(b)
By the assumption, $\lim_{x \to c} g(x) =L$; that is, for any $\epsilon > 0$, there exists a $\delta >0$ such that$$|x-c|< \delta \quad \Rightarrow \quad |g(x)-L|< \epsilon .$$The last inequality implies that$$- \epsilon < g(x)-L \quad \Rightarrow \quad L- \epsilon < g(x).$$Now, by letting $\epsilon = \frac{L}{2}$ (we can do this since $L>0$ and so $\frac{L}{2}>0$), it follows that$$L-\frac{L}{2}=\frac{L}{2} < g(x).$$Since the limit exists, there exists some $\delta _{0}$ corrosponding to $\epsilon = \frac{L}{2}$. By the above inequality we conclude that for all $x \neq c$ in the interval $(c-\delta _0 , c+ \delta _0)$, we have $g(x) > \frac{L}{2} >0$ (Please note that we excluded the point $x=c$ from this interval since in the definition of "limit" of a function $f$ at the point $c$, the function may not be defined at $c$ or may have any value, which does not have any relation to the existence of the limit).
