Solution:

(a) Let’s assume that any arbitrary real number is “a” then
$\lim f(x) \Rightarrow \lim f(a+h)$

So now,

$\lim _{h \rightarrow 0}(\sin a \cos h+\cos a \sin h+\cos a \cos h-\sin a \sin h)$

$=\sin a \cos 0+\cos a \sin 0+\cos a \cos 0-\sin a \sin 0$

$\{A s \cos 0=1$ and $\sin 0=0\}$

$=\sin a+\cos a=f(a)$ In the similar way,

$\lim _{x \rightarrow a^{-}} f(x)=f(a)$

$\lim _{x \rightarrow a^{-}} f(x)=f(a)=\lim _{x \rightarrow a^{\prime}} f(x)$

Since, $f(x)$ is continuous at $x=a$.

Since, “a” is an arbitrary real number, as a result, $f(x)=\sin x+\cos x$ is continuous.

(b) Let an arbitrary real number be “a” then $\lim _{x \rightarrow a^{-}} f(x) \Rightarrow \lim _{k \rightarrow 0} f(a+h)$

So now,

$\begin{array}{l}
\lim _{h \rightarrow 0} f(a+h)=\lim _{h \rightarrow 0} \sin (a+h)-\cos (a-h) \\
\Rightarrow \lim _{h \rightarrow 0}(\sin a \cos h+\cos a \sin h-\cos a \cos h-\sin a \sin h) \\
=\sin a \cos 0+\cos a \sin 0-\cos a \cos 0-\sin a \sin 0 \\
=\sin a+0-\cos a-0 \\
=\sin a-\cos a=f(a) \\\end{array}$
In the similar way,

$lim _{x \rightarrow a^{-}} f(x)=f(a)$
$\lim _{x \rightarrow a^{\circ}} f(x)=f(a)=\lim _{x \rightarrow a^{-}} f(x)$

As a result, $f(x)$ is continuous at $x=a$.

As, an arbitrary real number is “a”, as a result, $f(x)=\sin x-\cos x$ is continuous.