Solution:

(a) The provided function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$

For any real number, $k \neq-5$, we have

$\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{x^{2}-25}{x+5}=\lim _{x \rightarrow k} \frac{(x+5)(x-5)}{x+5}=\lim _{x \rightarrow k}(x-5)=k-5$

And $f(k)=\frac{(k+5)(k-5)}{k+5}=k-5$

Since, $\lim _{x \rightarrow k} f(x)=f(k)$, as a result, $f(x)$ is continuous at every point and it is a continuous function.

(b) The provided function is $f(x)=|x-5|$

For all real $x$ domain of $f(x)$ is real and infinite

Given here $f(x)=|x-5|$ is a modulus function.

Since we know that, every modulus function is continuous.

As a result, in its domain $\mathrm{R}$, $f$ is continuous .