Thank you all  for your answers!
If we replace the assigning by an usual substitution, the results remain identical. What's the difference?

restart;

V1:=Vector([a, b]);  V2:=<a, b>;  # Identical results

A:=subs({a=[1, 2],b=[3, 4]}, V1); B:=subs({a=[1, 2],b=[3, 4]}, V2); # Also identical results

whattype(A); whattype(B);

Since 1/1 = 1, then why just  1  not written?

Maybe should be  1/(1-exp(a*x))*erfc(a*x)  or  1/((1-exp(a*x))*erfc(a*x))  ?

@Alejandro Jakubi  Your way does not work if in solutions there is no the number  Pi . Here is more general way:

s := solve({x >= 0, 2*sin(Pi*x) = 1, x <= 2*Pi}, AllSolutions, Explicit);

seq({x = convert(rhs(op(s[i])), degrees)}, i = 1 .. nops([%]));

@acer 

At first I wrote  Diff  and Maple showed all the steps, then I replaced  Diff  by  diff  and Maple again showed all the steps. But if from the beginning written  diff, the error appears.

@oldstudent 

Maple 16

@Markiyan Hirnyk 

Thanks for the comment. Indication of editing done.

@oldstudent 

with(Student[Calculus1]):

A := diff(x^2*sin(x), x);

ShowSolution(A);

Another way to solve the problem, first create a matrix of the correct size, and then make the necessary substitutions or conversions.

x := [i $ i=1..25];

A:=Matrix(5, x);

eval(A, [seq(x[i]=convert(x[i], string), i=1..nops(x))]);

@bluehotel  Just copy the bold text below into Mathematica and press the key Enter:

Reduce [{-x3 * x5 /m - x3 * x5 - 2 * x1 * x5 + 2 * x1 == 0, 
x3 * x5 /m - 2 * x2 * x5 + 2 * x2 == 
0,-х3 + 2 * x5 - x3 * x5 + 1 - 3 * x1 - 3 * x2 + x4 == 
0, -2 * x4 - 2 * x4 * x5 == 0,-x3 * x5 ^ 2 / (x1 * m) - 4 * х5 ^ 2 + 4 * x5 == 
0}, {x1, x2, x3, x4, x5}] // TraditionalForm

@liumai   rhs((P(k))(t)[2])  is the right side of the second operand of the list  (P(k))(t)  for specific  k  and  t .

For 2 functions in the system replace  (k, t)->rhs((P(k))(t)[2])  by the list  [(k, t)->rhs((P(k))(t)[2]),(k, t)->rhs((P(k))(t)[4])]

@Carl Love  Your code in M 12 also works. I think  evala  command is crucial. I used  expand  command instead  and got the message "lost connection with the kernel"

@Alejandro Jakubi

1)  Thanks for the useful information regarding  parametric. 

2) You replaced the equation  sqrt(x-a) = x  with  the equation  x-a = x^2.  But  in the real domain, these equations are not equivalent.

@Carl Love  In my code the specific function  f  was used rather than arbitrary function as in your code. I think the questioner himself according this example will write the function that interests him.

@mahmood180  Your example for  tj=1:

HybrFunc(4, 3, 1):

for j from 0 to 2 do

seq('b'[i,j]=b[i,j](t), i=1..4);

od;

Download the text (not an image) of the code, so we can check its accuracy.