After 800+ views, I don't know why nobody has given upvote to this answer which explains the correct procedure.
For, $2^{nd}$ function, we can do like this :
$f(x) = x^2 - sinx \Rightarrow f'(x) = 2x - cosx$
Now, if $f'(x) = 0$ which means $2x - cosx = 0 \Rightarrow 2x = cosx$
Now, RHS $cosx$ lies between $-1$ and $+1$ [including both points] for all real $x$, So, LHS $2x$ should also lies between $-1$ and $+1$
So, $-1 \leq 2x \leq +1 \Rightarrow -1/2 \leq x \leq +1/2$
So, for the given question, if $f'(x) = 0$ then $0 \leq x \leq +1/2$
Same thing is true for other way means if $x$ is between $0$ and $1/2$ then $f'(x)=0$
So, $f'(x) = 0$ iff $x \in [0,1/2]$, Since, $f'(x)$ is not $>0$, So, it is not strictly increasing. Here, also, $f''(x) > 0$ in the given interval because $f''(x) = 2+sinx$. Since $sinx$ lies between $-1$ and $+1$ for all real $x$. So, $f''(x) = 2+sinx$ will lie between $1$ and $3$. Since, $f'(x)=0$ for interval $[0,1/2]$ and $f''(x)>0$, So, we can say $f(x)$ has atleast one local minima in the given interval. Hence, it can't be neither strictly increasing nor non-decreasing.
-- For the first function, if someone knows graph of $e^x$ but not $e^{-x}$. To make the graph of $f(-x)$ from $f(x)$, just reflect the $f(x)$ about y-axis means $y-axis$ will work as mirror. So, here reflect $e^x$ about y-axis.