Let $ f(x),\ g(x)$ be continuous functions defined in the range of $ 0\leq x\leq 1$.

(1) Find $ f(x)$ such that $ f(x)\equal{}\int_0^1 e^{x\plus{}t}f(t)\ dt$.

(2) Find $ g(x)$ such that $ g(x)\equal{}\int_0^1 e^{x\plus{}t}g(t)\ dt\plus{}x$.
### Part 1: Finding \( f(x) \)

1. Given the integral equation:
   \[
   f(x) = \int_0^1 e^{x+t} f(t) \, dt
   \]

2. Factor out \( e^x \) from the integrand:
   \[
   f(x) = e^x \int_0^1 e^t f(t) \, dt
   \]

3. Let \( C = \int_0^1 e^t f(t) \, dt \), which is a constant since the integral is independent of \( x \):
   \[
   f(x) = C e^x
   \]

4. Substitute \( f(x) = C e^x \) back into the original equation to determine \( C \):
   \[
   C e^x = \int_0^1 e^{x+t} C e^t \, dt
   \]

5. Simplify the right-hand side:
   \[
   C e^x = C e^x \int_0^1 e^{2t} \, dt
   \]

6. Evaluate the integral:
   \[
   \int_0^1 e^{2t} \, dt = \left[ \frac{e^{2t}}{2} \right]_0^1 = \frac{e^2 - 1}{2}
   \]

7. Equate both sides:
   \[
   C e^x = C e^x \cdot \frac{e^2 - 1}{2}
   \]

8. Since \( e^x \neq 0 \), we can divide both sides by \( e^x \):
   \[
   C = C \cdot \frac{e^2 - 1}{2}
   \]

9. Solve for \( C \):
   \[
   1 = \frac{e^2 - 1}{2} \implies 2 = e^2 - 1 \implies e^2 = 3 \implies C = 0
   \]

10. Therefore, \( f(x) = 0 \).

### Part 2: Finding \( g(x) \)

1. Given the integral equation:
   \[
   g(x) = \int_0^1 e^{x+t} g(t) \, dt + x
   \]

2. Assume a solution of the form \( g(x) = B e^x + x \):
   \[
   B e^x + x = \int_0^1 e^{x+t} (B e^t + t) \, dt + x
   \]

3. Split the integral:
   \[
   B e^x + x = \int_0^1 B e^{x+t} e^t \, dt + \int_0^1 e^{x+t} t \, dt + x
   \]

4. Evaluate the first integral:
   \[
   \int_0^1 B e^{x+2t} \, dt = B e^x \int_0^1 e^{2t} \, dt = B e^x \cdot \frac{e^2 - 1}{2}
   \]

5. Evaluate the second integral:
   \[
   \int_0^1 e^{x+t} t \, dt = e^x \int_0^1 e^t t \, dt
   \]

6. Use integration by parts for \( \int_0^1 e^t t \, dt \):
   \[
   \int_0^1 t e^t \, dt = \left[ t e^t \right]_0^1 - \int_0^1 e^t \, dt = e - \left[ e^t \right]_0^1 = e - (e - 1) = 1
   \]

7. Combine the results:
   \[
   B e^x + x = B e^x \cdot \frac{e^2 - 1}{2} + e^x + x
   \]

8. Simplify and equate coefficients:
   \[
   B e^x = B e^x \cdot \frac{e^2 - 1}{2} + e^x
   \]

9. Factor out \( e^x \):
   \[
   B = B \cdot \frac{e^2 - 1}{2} + 1
   \]

10. Solve for \( B \):
   \[
   B - B \cdot \frac{e^2 - 1}{2} = 1 \implies B \left(1 - \frac{e^2 - 1}{2}\right) = 1 \implies B \left(\frac{3 - e^2}{2}\right) = 1 \implies B = \frac{2}{3 - e^2}
   \]

11. Therefore, the solution is:
   \[
   g(x) = \frac{2}{3 - e^2} e^x + x
   \]

The final answer is \( \boxed{ f(x) = 0 } \) and \( g(x) = \frac{2}{3 - e^2} e^x + x \)