Method 1:
One-to-one: Let f(x) = f(y). Then 2x - 3 = 2y - 3; this implies that x = y. That is, f(x) = f(y) implies that x = y. Thus f is one-to-one.
Onto: Let y \(\in\) R. Let x \(={y+3 \over 2}.\)  Then \(f(x)=2\left( {y+3\over2} \right)-3=y.\)  Thus f is onto. This also can be proved by saying the following statement. The range of f is R (how?) which is equal to the co-domain and hence f is onto.
Inverse: Let y = 2x - 3. Then y + 3 = 2x and hence \(x={y+3\over 2}.\)  Thus \({f}^{-1}(y)={y+3\over2}\). By replacing y as x, we get \({f}^{-1}(x)={x+3\over 2}.\)
Method 2:
Let y = 2x − 3. Then \(x=\frac{y+3}{2}\). Let \(g(y)=\frac{y+3}{2}\)
Now
\((g \circ f)(x)=g(f(x))=g(2 x-3)=\frac{(2 x-3)+3}{2}=x .\)
\((f \circ g)(y)=f(g(y))=f\left(\frac{y+3}{2}\right)=2\left(\frac{y+3}{2}\right)-3=y\)
Thus, \(g \circ f=I_{X} \text { and } f \circ g=I_{Y}\)
This implies that f and g are bijections and inverses to each other. Hence f is a bijection and \(f^{-1}(y)=\frac{y+3}{2}\). Replacing y by x we get \(f^{-1}(x)=\frac{x+3}{2}\)

We know \(|x|=\begin{cases} -x\ \ if\ x\le0 \\ x\quad if\ x>0 \end{cases}\)
So, \(f(x)=\begin{cases} 2x-(-x)\quad if x \le 0\\ 2x-x\quad if\ x>0 \end{cases}\)
Thus, \(f(x)=\begin{cases}3x\quad ifx\le0\\x\quad if\ x>0 \end{cases}\)
Also, \(g(x)=\begin{cases} 2x+(-x)\quad if \ x\le 0\\2x+x\quad if\ x>0 \end{cases}\)
Thus, \(g(x)=\begin{cases} x\quad if x \le 0\\3x\quad if x >0 \end{cases}\)
Let \(x\le0.\) Then 
(f o g) (x) = f(g{x)) = f(x) = 3x.,
The last equality is taken because \(3x\le0\) whenever \(x\le0.\)
Let x > 0. Then
(f o g)(x) = f(g(x)) = f(3x) = 3x.
Thus (f o g)(x) = 3x for all x.

f(-3) = (-3)² - 3    \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\quad when\ x=-3 \right] \)
= 9 - 3 = 6
f(5) = 5² + 3(5)-2   \(\left[ \therefore f(x)={ x }^{ 2 }+3x-2\quad when\quad x=5 \right] \)
= 25 + 15 - 2
= 38
f(2) = 2² - 3
= 4 - 3 = 1   \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=2 \right] \)
f(-1) = (-1)² + (-1) -5   \(\left[ \therefore \ f(x)={ x }^{ 2 }+x-5\ when\ x=-1 \right] \)
= 1-1-5 = -5
f(0) = 0²-3 = -3  \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=0 \right] \)
\(\therefore\) f(-3) = 6, f(5) = 38, f(2) = 1, f(-1) = -5, f(0) = -3

(i) y = -x

Graph of y = - x is the reflection of the graph of y = x about the X - axis.
(ii) y = 2x

The graph of y = 2x compresses towards the Y-axis that is moves away from the X-axis since the multiplying factor is 2, which is greater than 1.
(iii) y = x + 1

The graph of y = x + 1, causes the shift to the upward for one unit.
(iv)  \(y={1\over 2}x+1\)

The graph of y = \(\frac{1}{2}\) x + 1, stretches towards the X-axis since the multiplying factor is \(\frac{1}{2}\) which is less than one and shift to the upward for one unit.
(v) ⇒ y = -2x - 3

The graph of y = -2x - 3, stretches towards the X-axis since the multiplying factor is - 2 which is less than one and causes the shifts to the downward for 3 units.

(i) \(y=-x^{ \left( \frac { 1 }{ 3 } \right) }\)

\(Let \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\)
\(Then \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\) is the reflection of the graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)  about the x-axis.
(ii) \(y=x^{1\over 3}+1\)

Let \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)
Then \(y=x^{ ^{ \frac { 1 }{ 3 } } }+1\) is the x graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the upward for one unit
(iii) \(y=x^{1\over 3}-1\)

Let \(y=x^{1\over 3}\)
Then \(y=x^{1\over 3}\)-1 is the graph of \(x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the downward for one unit.
(iv)  \(y=(x+1)^{1\over 3}\)

\(y=(x+1)^{1\over 3}\) causes the graph of \({x}^{\frac{1}{3}}\), shifts to the left for one unit.

(i) y = -x³

Let f(x) = x³
Since y = -f(x), this is the reflection of the graph off about the x-axis.
(ii)  y = x³+1

Let f(x) = x³
Since y = f(x) + 1, this is the graph of f(x) shifts to the upward for one unit.
(iii) y=x³-1

Let f(x) = x³
Since y = f(x) -1, this is the graph of (x) shifts to the downward for one unit.
(iv) y = (x + 1)³

Let f(x) = x³
y = (x + 1)³, causes the graph of f(x) shifts to the left for one unit.

If x < -3 or x > 3, then x² will be greater than 9 and hence 9 - x² will become negative which has no square root in R.
So x must lie on the interval [- 3, 3].
Also if \(x\ge-1\) or \(x\le 1,\) then x²-1 will become negative or zero. If it is negative, x² - 1 has no square root in R.  If it is zero, f is not defined. So, x must lie outside [- 1, 1]. 
That is x must lie on \(( -\infty,-1 ]\cup[1,\infty),\) Combining these two conditions, the largest possible domain for f is \([-3,3]\cap((-\infty, -1)\cup(1,\infty)).\)  That is \([-3,-1)\cup(1, 3].\)

Range of cosine function is -1 \(\le \)cos x \(\le \) 1
\(\Rightarrow\) -2 \(\le \) 2 cos x \(\le \) 2        (Multiplied by 2)
\(\Rightarrow\) -2 -1 \(\le \) 2 cos x -1 \(\le \) 2-1
\(\Rightarrow\) -3 \(\le \) 2 cos x-1 \(\le \) 1
\(\Rightarrow \frac { -1 }{ 3 } >\frac { 1 }{ 2cosx-1 } >\frac { 1 }{ 1 } \)
\(\Rightarrow \frac { -1 }{ 3 } f(x)>1\)
\(\therefore \ Range \) \(=\left(-\infty,-\frac{1}{3}\right] \cup[1, \infty)\)

(i) f( m) = f( n) \(\Rightarrow\) m² = n² \(\Rightarrow\) m = n since \(m,\ n\in N.\) Thus f is one-to-one. But, non-perfect square elements in the co-domain do not have pre-images and hence not onto.
(ii) Two different elements in the domain have same images and hence f is not one-to-one. Clearly the range of f is a proper subset of R. Thus it is not onto.