(i) No, any 5 measurements (elements) cannot determine a quadrilateral. Actually to construct a quadrilateral, we need a specific combination of measurements such as:
(a) Four sides and one diagonal
or
(b) Three sides and two diagonals
or
(c) Two adjacent sides and three angles
or
(d) Three sides and two included angles.
or
(e) Some special properties are given.
(ii) Let us draw a rough sketch of BATS as given.

Here, we cannot locate the points T and B without knowing the measurements ST and SB respectively. 
Thus, we cannot draw a parallelogram with the given measurements.
(iii) Let us draw a rough sketch of the (rhombus) quadrilateral ZEAL and mark the given measurements on it.

Yes, we can draw the required rhombus, because its all sides are equal to 3.5 em and a diagonal EL is known.
(iv) Let us draw the rough sketch of the quadrilateral PLAY and mark' its measurements.
No, this quadrilateral cannot be drawn.
\(\because\) Point P cannot be located. In \(\triangle\)LPY, sum of the lengths of PL and PY is less than LY, i.e., (2 cm + 3 cm) < 6 cm.

In cyclic quadrilateral,
\(\angle\)B = \(\angle\)D = 90°, [angle in a semi-circle]
\(\angle\)A = \(\angle\)C = 90° \(\Rightarrow \) \(\angle\)B + \(\angle\)D = 180°

\(\because \) \(\angle \)A = \(\angle \)B = \(\angle \)C= \(\angle \)D = 90°
Also, AB = BC = CD = DA
So, quadrilateral ABCD is a square.

Diagonals of a rectangle are equal.
\(\therefore\) AC = BD = 5 cm
Steps of construction
Step I Draw AB = 3 cm.
Step II Draw a ray BX such that \(\angle\) ABX = 90°.
Step III Draw an arc such that AC = 5 cm.
Step IV With B as centre, draw an arc of radius 5 cm and with C as centre. Draw another arc of radius 3 cm, which intersects first arc at a point, suppose D.
Step V Join CD and AD.

Hence ABCD is the required rectangle

Firstly, we draw a rough sketch of rectangle OKAY, which helps us in deciding steps of construction.

We know that, in a rectangle, opposite sides are equal and parallel and angles between two adjacent sides is 90°.
In rectangle OKAY,
OK = AY = 7 cm, KA = OY = 5 cm
and \(\angle \)OKA=\(\angle \)KOY = 90°
Steps of construction
Steps IDraw OK = 7 cm.
Step IIAt K, draw a ray KX making an \(\angle \)OKX = 90°
Step III Cut KA = 5 cm from ray KX.
Step IV With A as centre and radius 7 cm, draw an arc.
Step V With O as centre and radius 5 cm, draw another arc which intersects the arc drawn in Step IV at Y.
Step VI Join AY and OY.

Firstly, draw a rough sketch of quadrilateral MORE, which helps us in deciding steps of construction.
Steps of construction
Step I Draw MO = 6 cm.
Step II At O, draw a ray OX making \(\angle \)MOX = 105°.
Step III Cut OR = 4.5 cm on ray OX.
Step IV At M, draw a ray MY making \(\angle \)OMY =60°.
Step V At R, draw a ray RZ making \(\angle \)ORZ = 105°, which meets the ray MY at E.

PLAN is the required quadrilateral.

Firstly, draw a rough sketch of quadrilateral LIFT, which helps us in deciding steps of construction.

Steps of construction
Step I Draw LI = 4 cm
Step II With L as centre and radius 2.5 cm, draw an arc.
Step III With I as centre and radius 4 cm, draw another arc, which intersects the arc drawn in Step II at T.
Step IV With I as centre and radius 3 cm, draw an arc.
Step V With L as centre and radius 4.5 cm, draw another arc which intersects the arc drawn in Step IV at F.
Step Join IF, FT, TL, LF and IT.

GOLD is the required quadrilateral.

We know that, in a rhombus, all sides are of equal length.
Here, BE=4.5 cm
Firstly, draw a rough sketch of rhombus BEST, which helps us in deciding steps of construction.
So, BE = ES = ST = BT = 4.5 cm
Steps of construction
Step IDraw BE = 4.5 cm.
Step II With B as centre and radius 4.5 cm, draw an arc.
Step III With E as centre at E and radius 6 cm, draw another arc which intersects the arc in Step II at T
Step IV With E as centre and radius 4.5 cm, draw an arc on the side opposite to B with reference to ET
Step V With T as centre and radius 4.5 cm, draw another arc which intersects the arc drawn in step IV at S.
Step VI Join ES, ST, TB and TE.

Thus, BEST is the required rhombus. 

We know that, a quadrilateral has 8 elements, i.e. 4 sides and 4 angles. Firstly, draw a rough sketch of quadrilateral ABCD, which helps us in deciding of construction.

Steps of construction
Step IDraw AB = 4.5 cm.
Step II With A as centre and radius 7 em, draw an arc.
Step III With B as centre and radius 5.5 cm, draw another arc which intersects the arc drawn in Step II at C.
Step IV With A as centre and radius 6 cm, draw an arc on the side opposite to B with reference to AC.
Step V With C as centre and radius 4 cm, draw another arc which intersects the.arc drawn in Step IV at D.
Step VI Join BC, CD, DA and AC.

JUMP is the required quadrilateral.

Steps of construction
Step IDraw a line segment AB = 5 cm.

Step II At B, draw ray BX such that, \(\angle \)ABX = 90° .
Step III From ray BX, cut-off BC = 5 cm.

Step IV With centre at C, draw an arc above AB of radius 5 cm
Step V With centre at A and radius = 5 cm, draw another arc to intersect the previous arc at D.

Step VI Join DA and CD.

Thus, ABCD is a required square.

Initially, it appears that only two measurements are available. Actually, the figure is a special quadrilateral, so we have many more details with us.
We know that, in a rhombus diagonals bisect each other at right angle.
Steps of construction
Step I First, draw AC=8 cm.

Step II Now, construct its perpendicular bisector.

Step III Let them meet at O. Cut-off 3 cm lengths on either side of the drawn bisector. You now get B and D.

On joining AB, AB, BC and CD, we get the required rhombus.

First, we draw a rough sketch to visualise the quadrilateral, which is given below:

Steps of construction
Step I First draw LE = 4 cm and then, construct LE = 120° and cur the length EA = 4 cm on it.

Step II Make \(\angle \)EAY =90 ° on A

Step III Make\(\angle \)ELZ=60° and then get the intersection point \(\angle \)ELZ and \(\angle \) EAY, which gives a point P.

Thus, quadrilateral LEAP is constructed.

Let us draw a rough sketch to visualise the quadrilateral. See the following figure.

Steps of construction
Step I From the rough sketch, it is easy to see that a \(\Delta \)BCD can be constructed using SSS construction condition.
Draw the \(\Delta \)BCD.

Step II Now, we will locate a point A, which would be on the side opposite to C with reference to B. 
A is 5.5 cm away from D. So, with D as centre, draw an arc of radius 5.5 cm.

Step III A is 5 cm away from B. So, with B as centre, draw an arc of radius 5 cm.

Step IV A should be the intersection point of both the arcs drawn. Mark A and join BA and DA.

Hence, ABCD is the required quadrilateral.