Let $a,b,c$ be positive numbers such that $3a=b^3,\ 5a=c^2.$ Assume that a positive integer is limited to $d=1$ such that $a$ is divisible by $d^6.$

(1) Prove that $a$ is divisible by $3$ and $5.$
(2) Prove that the prime factor of $a$ are limited to $3$ and $5.$
(3) Find $a.$
Given the conditions:
\[ 3a = b^3 \]
\[ 5a = c^2 \]
and the assumption that \( a \) is divisible by \( d^6 \) where \( d = 1 \).

### Part 1: Prove that \( a \) is divisible by \( 3 \) and \( 5 \).

1. From the equation \( 3a = b^3 \), we can see that \( a = \frac{b^3}{3} \). For \( a \) to be an integer, \( b^3 \) must be divisible by \( 3 \). Since \( b \) is a positive integer, \( b \) itself must be divisible by \( 3 \). Let \( b = 3k \) for some integer \( k \). Then:
   \[
   b^3 = (3k)^3 = 27k^3
   \]
   Substituting back, we get:
   \[
   a = \frac{27k^3}{3} = 9k^3
   \]
   Thus, \( a \) is divisible by \( 3 \).

2. From the equation \( 5a = c^2 \), we can see that \( a = \frac{c^2}{5} \). For \( a \) to be an integer, \( c^2 \) must be divisible by \( 5 \). Since \( c \) is a positive integer, \( c \) itself must be divisible by \( 5 \). Let \( c = 5m \) for some integer \( m \). Then:
   \[
   c^2 = (5m)^2 = 25m^2
   \]
   Substituting back, we get:
   \[
   a = \frac{25m^2}{5} = 5m^2
   \]
   Thus, \( a \) is divisible by \( 5 \).

Since \( a \) is divisible by both \( 3 \) and \( 5 \), we conclude that \( a \) is divisible by \( 15 \).

### Part 2: Prove that the prime factors of \( a \) are limited to \( 3 \) and \( 5 \).

1. From the previous part, we have:
   \[
   a = 9k^3 \quad \text{and} \quad a = 5m^2
   \]
   Since \( a \) must satisfy both conditions simultaneously, we equate the two expressions:
   \[
   9k^3 = 5m^2
   \]
   This implies that \( a \) must be a common multiple of \( 9 \) and \( 5 \). The least common multiple of \( 9 \) and \( 5 \) is \( 45 \). Therefore, \( a \) must be of the form:
   \[
   a = 45n
   \]
   for some integer \( n \).

2. Since \( a \) is of the form \( 45n \), the prime factors of \( a \) are \( 3 \) and \( 5 \).

### Part 3: Find \( a \).

1. From the equation \( 9k^3 = 5m^2 \), we need to find integers \( k \) and \( m \) such that this equation holds. We rewrite it as:
   \[
   9k^3 = 5m^2
   \]
   This implies that \( k \) and \( m \) must be chosen such that both sides are equal. Since \( 9k^3 \) and \( 5m^2 \) must be equal, we need to find the smallest \( k \) and \( m \) that satisfy this condition.

2. By trial and error or by solving the equation, we find that the smallest values that satisfy this condition are \( k = 5 \) and \( m = 3 \). Substituting these values back, we get:
   \[
   a = 9k^3 = 9 \cdot 5^3 = 9 \cdot 125 = 1125
   \]
   \[
   a = 5m^2 = 5 \cdot 3^2 = 5 \cdot 9 = 45
   \]

Since both expressions must be equal, we find that the smallest value of \( a \) that satisfies both conditions is \( 1125 \).

The final answer is \( \boxed{1125} \).