If $f:X\to X$ is an epimorphism, which of these statements are true? 
$X$ is a linear space, $\dim X = n < \infty$.  $f:X\to X$ is epimorphism. Then:
  a. $f$ is monomorphism.
  b. there is exists base such that matrix of $f$ is diagonal.
  c. there is exists base such that matrix of $f$ is symmetric.  

a. is true, because:
Let $A$ be matrix of $f$. Since, $A$ is matrix of change base then this matrix is non-singular, so it is reversible.  Then $Ax=0\Leftrightarrow A^{-1}Ax=A^{-1}0 \Leftrightarrow x = 0$ so  $f$ is monomorphism.    
b. No idea,
c. No idea,  
What about a. ? Is it correct ? Can you help me with b., c. ? 
 A: For (a), your argument isn't really complete; I'd say you haven't fully justified how you know $A$ is a change-of-basis matrix.
But there's actually no need to talk about matrices at all, just use the rank-nullity theorem, the fact that $X$ is finite-dimensional, and the fact that


*

*$f$ is an epimorphism $\iff$ $\operatorname{im}(f)=X$

*$f$ is a monomorphism $\iff$ $\ker(f)=0$



For (b), remember that 


*

*$f$ is diagonalizable $\iff$ $\mathrm{algmult}(\lambda)=\mathrm{geomult}(\lambda)$ for every eigenvalue $\lambda$ of $f$


Take a look at a linear map like $f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$ from $\mathbb{R}^2$ to $\mathbb{R}^2$, which (you can check) is indeed an epimorphism.

The above map is also a good example to think about for part (c). Can there be a basis $\left\{\bigl\lbrack\begin{smallmatrix}a\\c\end{smallmatrix}\bigr\rbrack ,\bigl\lbrack\begin{smallmatrix}b\\d\end{smallmatrix}\bigr\rbrack\right\}$ of $\mathbb{R}^2$ in which the linear map given above is symmetric? Equivalently, can there be an invertible matrix $\bigl\lbrack\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr\rbrack$ (which remember, is equivalent to $ad-bc\neq 0$) such that
$$\begin{align*}
\begin{bmatrix}a & b\\ c & d\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}&=\begin{bmatrix}a & a+b\\ c & c+d\end{bmatrix}\begin{bmatrix}a & b\\ c & d\end{bmatrix}^{-1}\\[0.1in]
&=\frac{1}{ad-bc}\cdot\begin{bmatrix}a & a+b\\ c & c+d\end{bmatrix}\begin{bmatrix}\hphantom{-}d & -b\\ -c & \hphantom{-}a\end{bmatrix}\\[0.1in]
&=\frac{1}{ad-bc}\cdot\begin{bmatrix}ad-ac-bc & a^2\\ -c^2 & -bc+ac+ad\end{bmatrix}
\end{align*}$$
is symmetric? 
