a$.)$ The mutual reactance will be:

$x_{m1}=x_{11}-x_1=$

$x_{m2}=x_{22}-x_2=$

$x_{12}=\sqrt[\phantom{2}]{x_{m1}{x}_{m2}}=$

b$.)$ Taking as reference:

$V_2=\frac{2400}{{?}_0}V$

c$.)$ The secondary current module.

$|I_2|=\frac{S}{V_2}=$

d$.)$ The secondary current in phasor form.

$|)/{(}_C|=4,\frac{17}{{?}_{-}}Co{s}^{-1}0,9=$

e$.)$ The load impedance will be

$Z_L=\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_2}{\frac{?}{\text{b}\text{a}\text{r}}(I{)}_2}=\frac{\frac{2400}{{?}_0}}{4,\frac{17}{{?}_{-}}25,8}=$

f$.)$ Applying Kirchhoff's second law in the primary and secondary circuits.

$V_1=(r_1+j{x}_{11})I_1-j{x}_{12}{I}_2$

$r_1+j{x}_{11}=0,01+j20$

$V_1=(0,01+j20)I_1-j199,4*(3,75-j1,82)$

$0=(Z_L+r_2+j{x}_{22})-j{x}_{12}{I}_1$

$0=(518,17+j250,49+1+j2000)I_2-j199,4I_1$

In the above formula, substitute $,I_2$

And calculate.

a$.)$ The mutual reactance will be:

$x_{m1}=x_{11}-x_1=$

$x_{m2}=x_{22}-x_2=$

$x_{12}=\sqrt[\phantom{2}]{x_{m1}{x}_{m2}}=$

b$.)$ Taking as reference:

$V_2=\frac{2400}{{?}_0}V$

c$.)$ The secondary current module.

$|I_2|=\frac{S}{V_2}=$

d$.)$ The secondary current in phasor form.

$|)/{(}_C|=4,\frac{17}{{?}_{-}}Co{s}^{-1}0,9=$

e$.)$ The load impedance will be

$Z_L=\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_2}{\frac{?}{\text{b}\text{a}\text{r}}(I{)}_2}=\frac{\frac{2400}{{?}_0}}{4,\frac{17}{{?}_{-}}25,8}=$

f$.)$ Applying Kirchhoff's second law in the primary and secondary circuits.

$V_1=(r_1+j{x}_{11})I_1-j{x}_{12}{I}_2$

$r_1+j{x}_{11}=0,01+j20$

$V_1=(0,01+j20)I_1-j199,4*(3,75-j1,82)$

$0=(Z_L+r_2+j{x}_{22})-j{x}_{12}{I}_1$

$0=(518,17+j250,49+1+j2000)I_2-j199,4I_1$

In the above formula, substitute $,I_2$

And calculate.