The radius of the circular wire loop is $r=2.50,text{cm}=2.50times10^{-2},text{m}$, and its area is given as

$$
begin{align*}
A=r^2pi tag{1} ,.
end{align*}
$$

The magnitude of the magnetic field is $B=0.625,text{T}$, and the flux of the magnetic field through the circular loop is $Phi=1.00times10^{-4},text{T}cdottext{m}^2$. The expression for the flux is

$$
begin{align*}
Phi=BAcos{theta} tag{2},,
end{align*}
$$

where $theta$ is the angle between the normal of the loop and the direction of the magnetic field.

We want to calculate the value of $theta$.

Rearranging equation (2) we get

$$
begin{align*}
cos{theta}=frac{Phi}{BA},,
end{align*}
$$

which gives us

$$
begin{align*}
theta=arccos{left(frac{Phi}{BA}right)},.
end{align*}
$$

Putting the expression (1) into the last equation we obtain

$$
begin{align*}
theta&=arccos{left(frac{Phi}{Br^2pi}right)} tag{3}\
&=arccos{left(frac{ 1.00times10^{-4},text{T}cdottext{m}^2 }{(0.625,text{T} )(2.50times10^{-2},text{m})^2pi}right)}\\
&boxed{theta=85.3^{circ}},,
end{align*}
$$

where we changed in the values of $Phi$, $B$ and $r$ in line (3).

The flux of the magnetic field through the circular loop is $Phi=4.6times10^{-4},text{T}cdottext{m}^2$. The normal of the loop makes an angle of $theta=45^{circ}$ with the direction of the magnetic field, and the magnitude of the magnetic field is $B=0.50,text{T}$. The expression for the flux is

$$
begin{align*}
Phi=BAcos{theta} tag{1},,
end{align*}
$$

where $A$ is the area of the loop.

We want to calculate the value of $A$.

Rearranging equation (1) we get

$$
begin{align*}
A&=frac{Phi}{Bcos{theta}} tag{2} \
&=frac{4.6times10^{-4},text{T}cdottext{m}^2}{(0.50,text{T})cos{45^{circ}}} \\
&boxed{A=1.3times10^{-3},text{m}^2=13,text{cm}^2} ,,
end{align*}
$$

where we changed in the values of $Phi$, $B$ and $theta$ in line (2).

The width of the rectangular wire loop is $a=3.2,text{cm}=3.2times10^{-2},text{m}$, and its length is $b=5.1,text{cm}=5.1times10^{-2},text{m}$. The area of the loop is given as

$$
begin{align*}
A=ab tag{1} ,.
end{align*}
$$

The angle between the normal of the loop and the direction of the magnetic field is $theta=32^{circ}$, and the flux of the magnetic field through the loop is $Phi=2.2times10^{-3},text{T}cdottext{m}^2$. The expression for the flux is

$$
begin{align*}
Phi=BAcos{theta} tag{2},,
end{align*}
$$

where $B$ is the magnitude of the magnetic field.

We want to calculate the value of $B$.

Rearranging equation (2) we get

$$
begin{align*}
B=frac{Phi}{Acos{theta}},,
end{align*}
$$

Putting the expression (1) into the last equation we obtain

$$
begin{align*}
B&=frac{Phi}{abcos{theta}} tag{3}\
&=frac{2.2times10^{-3},text{T}cdottext{m}^2}{(3.2times10^{-2},text{m})(5.1times10^{-2},text{m})cos{32^{circ}}} \\
&boxed{B=1.6,text{T}},,
end{align*}
$$

where we changed in the values of $Phi$, $a$, $b$ and $theta$ in line (3).